I have a fairly large set of 2D points (~20000) in a set, and for each point in the x-y plane want to determine which point from the set is closest. (Actually, the points are of different types, and I just want to know which type is closest. And the x-y plane is a bitmap, say 640x480.)
From this answer to the question "All k nearest neighbors in 2D, C++" I got the idea to make a grid. I created n*m C++ vectors and put the points in the vector, depending on which bin it falls into. The idea is that you only have to check the distance of the points in the bin, instead of all points. If there is no point in the bin, you continue with the adjacent bins in a spiralling manner.
Unfortunately, I only read Oli Charlesworth's comment afterwards:
Not just adjacent, unfortunately (consider that points in the cell two
to the east may be closer than points in the cell directly north-east,
for instance; this problem gets much worse in higher dimensions).
Also, what if the neighbouring cells happen to have less than 10
points in them? In practice, you will need to "spiral out".
Fortunately, I already had the spiraling code figured out (a nice C++ version here, and there are other versions in the same question). But I'm still left with the problem:
If I find a hit in a cell, there could be a closer hit in an adjacent cell (yellow is my probe, red is the wrong choice, green the actual closest point):
If I find a hit in an adjacent cell, there could be a hit in a cell 2 steps away, as Oli Charlesworth remarked:
But even worse, if I find a hit in a cell two steps away, there could still be a closer hit in a hit three steps away! That means I'd have to consider all cells with dx,dy= -3...3, or 49 cells!
Now, in practice this won't happen often, because I can choose my bin size so the cells are filled enough. Still, I'd like to have a correct result, without iterating over all points.
So how do I find out when to stop "spiralling" or searching? I heard there is an approach with multiple overlapping grids, but I didn't quite understand it. Is it possible to salvage this grid technique?
Since the dimensions of your bitmap are not large and you want to calculate the closest point for every (x,y), you can use dynamic programming.
Let V[i][j] be the distance from (i,j) to the closest point in the set, but considering only the points in the set that are in the "rectangle" [(1, 1), (i, j)].
Then V[i][j] = 0 if there is a point in (i, j), or V[i][j] = min(V[i'][j'] + dist((i, j), (i', j'))) where (i', j') is one of the three neighbours of (i,j):
i.e.
(i - 1, j)
(i, j - 1)
(i - 1, j - 1)
This gives you the minimum distance, but only for the "upper left" rectangle. We do the same for the "upper right", "lower left", and "lower right" orientations, and then take the minimum.
The complexity is O(size of the plane), which is optimal.
For you task usually a Point Quadtree is used, especially when the points are not evenly distributed.
To save main memory you als can use a PM or PMR-Quadtree which uses buckets.
You search in your cell and in worst case all quad cells surounding the cell.
You can also use a k-d tree.
A solution im trying
First make a grid such that you have an average of say 1 (more if you want larger scan) points per box.
Select the center box. Continue selecting neighbor boxes in a circular manner until you find at least one neighbor. At this point you can have 1 or 9 or so on boxes selected
Select one more layer of adjacent boxes
Now you have a fairly small list of points, usually not more than 10 which you can punch into the distance formula to find the nearest neighbor.
Since you have on average 1 points per box, you will mostly be selecting 9 boxes and comparing 9 distances. Can adjust grid size according to your dataset properties to achieve better results.
Also, if your data has a lot of variance, you can try 2 levels of grid (or even more) so if selection works and returns more than 50 points in a single query, start a next grid search with a grid 1/10th the size ...
One solution would be to construct multiple partitionings with different grid sizes.
Assume you create partitions at levels 1,2,4,8,..
Now, search for a point in grid size 1 (you are basically searching in 9 squares). If there is a point in the search area and if distance to that point is less than 1, stop. Otherwise move on to the next grid size.
The number of grids you need to construct is about twice as compared to creating just one level of partitioning.
Related
I have non empty Set of points scattered on plane, they are given by their coordinates.
Problem is to quickly reply such queries:
Give me the point from your set which is nearest to the point A(x, y)
My current solution pseudocode
query( given_point )
{
nearest_point = any point from Set
for each point in Set
if dist(point, query_point) < dist(nearest_point, given_point)
nearest_point = point
return nearest_point
}
But this algorithm is very slow with complexity is O(N).
The question is, is there any data structure or tricky algorithms with precalculations which will dramatically reduce time complexity? I need at least O(log N)
Update
By distance I mean Euclidean distance
You can get O(log N) time using a kd-tree. This is like a binary search tree, except that it splits points first on the x-dimension, then the y-dimension, then the x-dimension again, and so on.
If your points are homogeneously distributed, you can achieve O(1) look-up by binning the points into evenly-sized boxes and then searching the box in which the query point falls and its eight neighbouring boxes.
It would be difficult to make an efficient solution from Voronoi diagrams since this requires that you solve the problem of figuring out which Voronoi cell the query point falls in. Much of the time this involves building an R*-tree to query the bounding boxes of the Voronoi cells (in O(log N) time) and then performing point-in-polygon checks (O(p) in the number of points in the polygon's perimeter).
You can divide your grid in subsections:
Depending on the number of points and grid size, you choose a useful division. Let's assume a screen of 1000x1000 pixels, filled with random points, evenly distributed over the surface.
You may divide the screen into 10x10 sections and make a map (roughX, roughY)->(List ((x, y), ...). For a certain point, you may lookup all points in the same cell and - since the point may be closer to points of the neighbor cell than to an extreme point in the same cell, the surrounding cells, maybe even 2 cells away. This would reduce the searching scope to 16 cells.
If you don't find a point in the same cell/layer, expand the search to next layer.
If you happen to find the next neighbor in one of the next layers, you have to expand the searching scope to an additional layer for each layer. If there are too many points, choose a finer grid. If there are to few points, choose a bigger grid. Note, that the two green circles, connected to the red with a line, have the same distance to the red one, but one is in layer 0 (same cell) but the other layer 2 (next of next cell).
Without preprocessing you definitely need to spend O(N), as you must look at every point before return the closest.
You can look here Nearest neighbor search for how to approach this problem.
N.B: there's a major edit at the bottom of the question - check it out
Question
Say I have a set of points:
I want to find the point with the most points surrounding it, within radius (ie a circle) or within (ie a square) of the point for 2 dimensions. I'll refer to it as the densest point function.
For the diagrams in this question, I'll represent the surrounding region as circles. In the image above, the middle point's surrounding region is shown in green. This middle point has the most surrounding points of all the points within radius and would be returned by the densest point function.
What I've tried
A viable way to solve this problem would be to use a range searching solution; this answer explains further and that it has " worst-case time". Using this, I could get the number of points surrounding each point and choose the point with largest surrounding point count.
However, if the points were extremely densely packed (in the order of a million), as such:
then each of these million points () would need to have a range search performed. The worst-case time , where is the number of points returned in the range, is true for the following point tree types:
kd-trees of two dimensions (which are actually slightly worse, at ),
2d-range trees,
Quadtrees, which have a worst-case time of
So, for a group of points within radius of all points within the group, it gives complexity of for each point. This yields over a trillion operations!
Any ideas on a more efficient, precise way of achieving this, so that I could find the point with the most surrounding points for a group of points, and in a reasonable time (preferably or less)?
EDIT
Turns out that the method above is correct! I just need help implementing it.
(Semi-)Solution
If I use a 2d-range tree:
A range reporting query costs , for returned points,
For a range tree with fractional cascading (also known as layered range trees) the complexity is ,
For 2 dimensions, that is ,
Furthermore, if I perform a range counting query (i.e., I do not report each point), then it costs .
I'd perform this on every point - yielding the complexity I desired!
Problem
However, I cannot figure out how to write the code for a counting query for a 2d layered range tree.
I've found a great resource (from page 113 onwards) about range trees, including 2d-range tree psuedocode. But I can't figure out how to introduce fractional cascading, nor how to correctly implement the counting query so that it is of O(log n) complexity.
I've also found two range tree implementations here and here in Java, and one in C++ here, although I'm not sure this uses fractional cascading as it states above the countInRange method that
It returns the number of such points in worst case
* O(log(n)^d) time. It can also return the points that are in the rectangle in worst case
* O(log(n)^d + k) time where k is the number of points that lie in the rectangle.
which suggests to me it does not apply fractional cascading.
Refined question
To answer the question above therefore, all I need to know is if there are any libraries with 2d-range trees with fractional cascading that have a range counting query of complexity so I don't go reinventing any wheels, or can you help me to write/modify the resources above to perform a query of that complexity?
Also not complaining if you can provide me with any other methods to achieve a range counting query of 2d points in in any other way!
I suggest using plane sweep algorithm. This allows one-dimensional range queries instead of 2-d queries. (Which is more efficient, simpler, and in case of square neighborhood does not require fractional cascading):
Sort points by Y-coordinate to array S.
Advance 3 pointers to array S: one (C) for currently inspected (center) point; other one, A (a little bit ahead) for nearest point at distance > R below C; and the last one, B (a little bit behind) for farthest point at distance < R above it.
Insert points pointed by A to Order statistic tree (ordered by coordinate X) and remove points pointed by B from this tree. Use this tree to find points at distance R to the left/right from C and use difference of these points' positions in the tree to get number of points in square area around C.
Use results of previous step to select "most surrounded" point.
This algorithm could be optimized if you rotate points (or just exchange X-Y coordinates) so that width of the occupied area is not larger than its height. Also you could cut points into vertical slices (with R-sized overlap) and process slices separately - if there are too many elements in the tree so that it does not fit in CPU cache (which is unlikely for only 1 million points). This algorithm (optimized or not) has time complexity O(n log n).
For circular neighborhood (if R is not too large and points are evenly distributed) you could approximate circle with several rectangles:
In this case step 2 of the algorithm should use more pointers to allow insertion/removal to/from several trees. And on step 3 you should do a linear search near points at proper distance (<=R) to distinguish points inside the circle from the points outside it.
Other way to deal with circular neighborhood is to approximate circle with rectangles of equal height (but here circle should be split into more pieces). This results in much simpler algorithm (where sorted arrays are used instead of order statistic trees):
Cut area occupied by points into horizontal slices, sort slices by Y, then sort points inside slices by X.
For each point in each slice, assume it to be a "center" point and do step 3.
For each nearby slice use binary search to find points with Euclidean distance close to R, then use linear search to tell "inside" points from "outside" ones. Stop linear search where the slice is completely inside the circle, and count remaining points by difference of positions in the array.
Use results of previous step to select "most surrounded" point.
This algorithm allows optimizations mentioned earlier as well as fractional cascading.
I would start by creating something like a https://en.wikipedia.org/wiki/K-d_tree, where you have a tree with points at the leaves and each node information about its descendants. At each node I would keep a count of the number of descendants, and a bounding box enclosing those descendants.
Now for each point I would recursively search the tree. At each node I visit, either all of the bounding box is within R of the current point, all of the bounding box is more than R away from the current point, or some of it is inside R and some outside R. In the first case I can use the count of the number of descendants of the current node to increase the count of points within R of the current point and return up one level of the recursion. In the second case I can simply return up one level of the recursion without incrementing anything. It is only in the intermediate case that I need to continue recursing down the tree.
So I can work out for each point the number of neighbours within R without checking every other point, and pick the point with the highest count.
If the points are spread out evenly then I think you will end up constructing a k-d tree where the lower levels are close to a regular grid, and I think if the grid is of size A x A then in the worst case R is large enough so that its boundary is a circle that intersects O(A) low level cells, so I think that if you have O(n) points you could expect this to cost about O(n * sqrt(n)).
You can speed up whatever algorithm you use by preprocessing your data in O(n) time to estimate the number of neighbouring points.
For a circle of radius R, create a grid whose cells have dimension R in both the x- and y-directions. For each point, determine to which cell it belongs. For a given cell c this test is easy:
c.x<=p.x && p.x<=c.x+R && c.y<=p.y && p.y<=c.y+R
(You may want to think deeply about whether a closed or half-open interval is correct.)
If you have relatively dense/homogeneous coverage, then you can use an array to store the values. If coverage is sparse/heterogeneous, you may wish to use a hashmap.
Now, consider a point on the grid. The extremal locations of a point within a cell are as indicated:
Points at the corners of the cell can only be neighbours with points in four cells. Points along an edge can be neighbours with points in six cells. Points not on an edge are neighbours with points in 7-9 cells. Since it's rare for a point to fall exactly on a corner or edge, we assume that any point in the focal cell is neighbours with the points in all 8 surrounding cells.
So, if a point p is in a cell (x,y), N[p] identifies the number of neighbours of p within radius R, and Np[y][x] denotes the number of points in cell (x,y), then N[p] is given by:
N[p] = Np[y][x]+
Np[y][x-1]+
Np[y-1][x-1]+
Np[y-1][x]+
Np[y-1][x+1]+
Np[y][x+1]+
Np[y+1][x+1]+
Np[y+1][x]+
Np[y+1][x-1]
Once we have the number of neighbours estimated for each point, we can heapify that data structure into a maxheap in O(n) time (with, e.g. make_heap). The structure is now a priority-queue and we can pull points off in O(log n) time per query ordered by their estimated number of neighbours.
Do this for the first point and use a O(log n + k) circle search (or some more clever algorithm) to determine the actual number of neighbours the point has. Make a note of this point in a variable best_found and update its N[p] value.
Peek at the top of the heap. If the estimated number of neighbours is less than N[best_found] then we are done. Otherwise, repeat the above operation.
To improve estimates you could use a finer grid, like so:
along with some clever sliding window techniques to reduce the amount of processing required (see, for instance, this answer for rectangular cases - for circular windows you should probably use a collection of FIFO queues). To increase security you can randomize the origin of the grid.
Considering again the example you posed:
It's clear that this heuristic has the potential to save considerable time: with the above grid, only a single expensive check would need to be performed in order to prove that the middle point has the most neighbours. Again, a higher-resolution grid will improve the estimates and decrease the number of expensive checks which need to be made.
You could, and should, use a similar bounding technique in conjunction with mcdowella's answers; however, his answer does not provide a good place to start looking, so it is possible to spend a lot of time exploring low-value points.
Suppose we have n points in a bounded region of the plane. The problem is to divide it in 4 regions (with a horizontal and a vertical line) such that the sum of a metric in each region is minimized.
The metric can be for example, the sum of the distances between the points in each region ; or any other measure about the spreadness of the points. See the figure below.
I don't know if any clustering algorithm might help me tackle this problem, or if for instance it can be formulated as a simple optimization problem. Where the decision variables are the "axes".
I believe this can be formulated as a MIP (Mixed Integer Programming) problem.
Lets introduce 4 quadrants A,B,C,D. A is right,upper, B is right,lower, etc. Then define a binary variable
delta(i,k) = 1 if point i is in quadrant k
0 otherwise
and continuous variables
Lx, Ly : coordinates of the lines
Obviously we have:
sum(k, delta(i,k)) = 1
xlo <= Lx <= xup
ylo <= Ly <= yup
where xlo,xup are the minimum and maximum x-coordinate. Next we need to implement implications like:
delta(i,'A') = 1 ==> x(i)>=Lx and y(i)>=Ly
delta(i,'B') = 1 ==> x(i)>=Lx and y(i)<=Ly
delta(i,'C') = 1 ==> x(i)<=Lx and y(i)<=Ly
delta(i,'D') = 1 ==> x(i)<=Lx and y(i)>=Ly
These can be handled by so-called indicator constraints or written as linear inequalities, e.g.
x(i) <= Lx + (delta(i,'A')+delta(i,'B'))*(xup-xlo)
Similar for the others. Finally the objective is
min sum((i,j,k), delta(i,k)*delta(j,k)*d(i,j))
where d(i,j) is the distance between points i and j. This objective can be linearized as well.
After applying a few tricks, I could prove global optimality for 100 random points in about 40 seconds using Cplex. This approach is not really suited for large datasets (the computation time quickly increases when the number of points becomes large).
I suspect this cannot be shoe-horned into a convex problem. Also I am not sure this objective is really what you want. It will try to make all clusters about the same size (adding a point to a large cluster introduces lots of distances to be added to the objective; adding a point to a small cluster is cheap). May be an average distance for each cluster is a better measure (but that makes the linearization more difficult).
Note - probably incorrect. I will try and add another answer
The one dimensional version of minimising sums of squares of differences is convex. If you start with the line at the far left and move it to the right, each point crossed by the line stops accumulating differences with the points to its right and starts accumulating differences to the points to its left. As you follow this the differences to the left increase and the differences to the right decrease, so you get a monotonic decrease, possibly a single point that can be on either side of the line, and then a monotonic increase.
I believe that the one dimensional problem of clustering points on a line is convex, but I no longer believe that the problem of drawing a single vertical line in the best position is convex. I worry about sets of points that vary in y co-ordinate so that the left hand points are mostly high up, the right hand points are mostly low down, and the intermediate points alternate between high up and low down. If this is not convex, the part of the answer that tries to extend to two dimensions fails.
So for the one dimensional version of the problem you can pick any point and work out in time O(n) whether that point should be to the left or right of the best dividing line. So by binary chop you can find the best line in time O(n log n).
I don't know whether the two dimensional version is convex or not but you can try all possible positions for the horizontal line and, for each position, solve for the position of the vertical line using a similar approach as for the one dimensional problem (now you have the sum of two convex functions to worry about, but this is still convex, so that's OK). Therefore you solve at most O(n) one-dimensional problems, giving cost O(n^2 log n).
If the points aren't very strangely distributed, I would expect that you could save a lot of time by using the solution of the one dimensional problem at the previous iteration as a first estimate of the position of solution for the next iteration. Given a starting point x, you find out if this is to the left or right of the solution. If it is to the left of the solution, go 1, 2, 4, 8... steps away to find a point to the right of the solution and then run binary chop. Hopefully this two-stage chop is faster than starting a binary chop of the whole array from scratch.
Here's another attempt. Lay out a grid so that, except in the case of ties, each point is the only point in its column and the only point in its row. Assuming no ties in any direction, this grid has N rows, N columns, and N^2 cells. If there are ties the grid is smaller, which makes life easier.
Separating the cells with a horizontal and vertical line is pretty much picking out a cell of the grid and saying that cell is the cell just above and just to the right of where the lines cross, so there are roughly O(N^2) possible such divisions, and we can calculate the metric for each such division. I claim that when the metric is the sum of the squares of distances between points in a cluster the cost of this is pretty much a constant factor in an O(N^2) problem, so the whole cost of checking every possibility is O(N^2).
The metric within a rectangle formed by the dividing lines is SUM_i,j[ (X_i - X_j)^2 + (Y_i-Y_j)^2]. We can calculate the X contributions and the Y contributions separately. If you do some algebra (which is easier if you first subtract a constant so that everything sums to zero) you will find that the metric contribution from a co-ordinate is linear in the variance of that co-ordinate. So we want to calculate the variances of the X and Y co-ordinates within the rectangles formed by each division. https://en.wikipedia.org/wiki/Algebraic_formula_for_the_variance gives us an identity which tells us that we can work out the variance given SUM_i Xi and SUM_i Xi^2 for each rectangle (and the corresponding information for the y co-ordinate). This calculation can be inaccurate due to floating point rounding error, but I am going to ignore that here.
Given a value associated with each cell of a grid, we want to make it easy to work out the sum of those values within rectangles. We can create partial sums along each row, transforming 0 1 2 3 4 5 into 0 1 3 6 10 15, so that each cell in a row contains the sum of all the cells to its left and itself. If we take these values and do partial sums up each column, we have just worked out, for each cell, the sum of the rectangle whose top right corner lies in that cell and which extends to the bottom and left sides of the grid. These calculated values at the far right column give us the sum for all the cells on the same level as that cell and below it. If we subtract off the rectangles we know how to calculate we can find the value of a rectangle which lies at the right hand side of the grid and the bottom of the grid. Similar subtractions allow us to work out first the value of the rectangles to the left and right of any vertical line we choose, and then to complete our set of four rectangles formed by two lines crossing by any cell in the grid. The expensive part of this is working out the partial sums, but we only have to do that once, and it costs only O(N^2). The subtractions and lookups used to work out any particular metric have only a constant cost. We have to do one for each of O(N^2) cells, but that is still only O(N^2).
(So we can find the best clustering in O(N^2) time by working out the metrics associated with all possible clusterings in O(N^2) time and choosing the best).
Given a starting square (y, x) on a 2d grid, I want to find the closest empty square to it. (Note: The 4 squares adjacent to the starting square should be considered closer than the 4 diagonal squares nearest it.)
The following image shows the order that I need to check the following cells on this grid:
The grid is bounded but can be quite large. In practice, the starting coordinate will be randomly located around the grid. (So, I don't think it's too important to worry about coordinates outside the bounds of the grid....)
What algorithm can I use to iterate around the circle in this manner?
A simple breadth-first search will do it. Push each neighbour to be examined onto a heap, prioritised by distance. You can probably get away with manhattan distance (dx + dy), but if not just use squared radial distance (dx2 + dy2). Whenever you pop an item it'll be the nearest. If it's empty, you found it. Otherwise push its neighbours onto the heap and keep popping.
I would probably use the square radial distance and only add adjacent squares (not diagonals). The diagonals will be considered later because they are immediately adjacent to other squares. You do need a way to keep track of which squares have already been considered so you don't add them again. There must be a clever dynamic-programming way of tracking this without having to clear a large grid of booleans each time you search... But saying that, a large grid of booleans will do quite nicely.
It can be solved with BFS (Breadth First Search). We have to process each square twice. First time we visit the still unvisited squares those share an edge with the current square and next time we visit the squares those share at least a point with the current square(Diagonally adjacent squares)
We can use two different queues to ensure that before processing a square 2nd time all squares with equal distance from the source to the current square have been processed at least once. :-)
Average Runtime: O(V*8) => O(V). Where V is the number of square inside the grid
If the content of the grid changes often, use methods described in previous answers, that is bread-first search.
If the content of your grid does changes seldom AND Manhattan distance is ok for your application, my advise is to compute the distance transform of the binarized grid (0 if empty, 1 else) distance transform is very simple for Manhattan distance, way more complicated for euclidian distance). This step can be done at a cost of 2*N*M (number of elements of the grid). Then, for each request, you can visit the neighborhood in a very straightforward manner, that is follow the path of min distance starting from the starting cell (like some gradient descent), it will stop at the nearest empty cell. May be really faster to search with this algorithm, as you don't look in the wrong way for an empty cell more than 1 cell far.
What is the most efficient way to randomly fill a space with as many non-overlapping shapes? In my specific case, I'm filling a circle with circles. I'm randomly placing circles until either a certain percentage of the outer circle is filled OR a certain number of placements have failed (i.e. were placed in a position that overlapped an existing circle). This is pretty slow, and often leaves empty spaces unless I allow a huge number of failures.
So, is there some other type of filling algorithm I can use to quickly fill as much space as possible, but still look random?
Issue you are running into
You are running into the Coupon collector's problem because you are using a technique of Rejection sampling.
You are also making strong assumptions about what a "random filling" is. Your algorithm will leave large gaps between circles; is this what you mean by "random"? Nevertheless it is a perfectly valid definition, and I approve of it.
Solution
To adapt your current "random filling" to avoid the rejection sampling coupon-collector's issue, merely divide the space you are filling into a grid. For example if your circles are of radius 1, divide the larger circle into a grid of 1/sqrt(2)-width blocks. When it becomes "impossible" to fill a gridbox, ignore that gridbox when you pick new points. Problem solved!
Possible dangers
You have to be careful how you code this however! Possible dangers:
If you do something like if (random point in invalid grid){ generateAnotherPoint() } then you ignore the benefit / core idea of this optimization.
If you do something like pickARandomValidGridbox() then you will slightly reduce the probability of making circles near the edge of the larger circle (though this may be fine if you're doing this for a graphics art project and not for a scientific or mathematical project); however if you make the grid size 1/sqrt(2) times the radius of the circle, you will not run into this problem because it will be impossible to draw blocks at the edge of the large circle, and thus you can ignore all gridboxes at the edge.
Implementation
Thus the generalization of your method to avoid the coupon-collector's problem is as follows:
Inputs: large circle coordinates/radius(R), small circle radius(r)
Output: set of coordinates of all the small circles
Algorithm:
divide your LargeCircle into a grid of r/sqrt(2)
ValidBoxes = {set of all gridboxes that lie entirely within LargeCircle}
SmallCircles = {empty set}
until ValidBoxes is empty:
pick a random gridbox Box from ValidBoxes
pick a random point inside Box to be center of small circle C
check neighboring gridboxes for other circles which may overlap*
if there is no overlap:
add C to SmallCircles
remove the box from ValidBoxes # possible because grid is small
else if there is an overlap:
increase the Box.failcount
if Box.failcount > MAX_PERGRIDBOX_FAIL_COUNT:
remove the box from ValidBoxes
return SmallCircles
(*) This step is also an important optimization, which I can only assume you do not already have. Without it, your doesThisCircleOverlapAnother(...) function is incredibly inefficient at O(N) per query, which will make filling in circles nearly impossible for large ratios R>>r.
This is the exact generalization of your algorithm to avoid the slowness, while still retaining the elegant randomness of it.
Generalization to larger irregular features
edit: Since you've commented that this is for a game and you are interested in irregular shapes, you can generalize this as follows. For any small irregular shape, enclose it in a circle that represent how far you want it to be from things. Your grid can be the size of the smallest terrain feature. Larger features can encompass 1x2 or 2x2 or 3x2 or 3x3 etc. contiguous blocks. Note that many games with features that span large distances (mountains) and small distances (torches) often require grids which are recursively split (i.e. some blocks are split into further 2x2 or 2x2x2 subblocks), generating a tree structure. This structure with extensive bookkeeping will allow you to randomly place the contiguous blocks, however it requires a lot of coding. What you can do however is use the circle-grid algorithm to place the larger features first (when there's lot of space to work with on the map and you can just check adjacent gridboxes for a collection without running into the coupon-collector's problem), then place the smaller features. If you can place your features in this order, this requires almost no extra coding besides checking neighboring gridboxes for collisions when you place a 1x2/3x3/etc. group.
One way to do this that produces interesting looking results is
create an empty NxM grid
create an empty has-open-neighbors set
for i = 1 to NumberOfRegions
pick a random point in the grid
assign that grid point a (terrain) type
add the point to the has-open-neighbors set
while has-open-neighbors is not empty
foreach point in has-open-neighbors
get neighbor-points as the immediate neighbors of point
that don't have an assigned terrain type in the grid
if none
remove point from has-open-neighbors
else
pick a random neighbor-point from neighbor-points
assign its grid location the same (terrain) type as point
add neighbor-point to the has-open-neighbors set
When done, has-open-neighbors will be empty and the grid will have been populated with at most NumberOfRegions regions (some regions with the same terrain type may be adjacent and so will combine to form a single region).
Sample output using this algorithm with 30 points, 14 terrain types, and a 200x200 pixel world:
Edit: tried to clarify the algorithm.
How about using a 2-step process:
Choose a bunch of n points randomly -- these will become the centres of the circles.
Determine the radii of these circles so that they do not overlap.
For step 2, for each circle centre you need to know the distance to its nearest neighbour. (This can be computed for all points in O(n^2) time using brute force, although it may be that faster algorithms exist for points in the plane.) Then simply divide that distance by 2 to get a safe radius. (You can also shrink it further, either by a fixed amount or by an amount proportional to the radius, to ensure that no circles will be touching.)
To see that this works, consider any point p and its nearest neighbour q, which is some distance d from p. If p is also q's nearest neighbour, then both points will get circles with radius d/2, which will therefore be touching; OTOH, if q has a different nearest neighbour, it must be at distance d' < d, so the circle centred at q will be even smaller. So either way, the 2 circles will not overlap.
My idea would be to start out with a compact grid layout. Then take each circle and perturb it in some random direction. The distance in which you perturb it can also be chosen at random (just make sure that the distance doesn't make it overlap another circle).
This is just an idea and I'm sure there are a number of ways you could modify it and improve upon it.