What do i mean by path density? Consider a simple 4-connected grid with two points two spaces apart
0 0 0 0 0 Allowed 0 1 0
0 A 0 B 0 Moves => 1 0 1
0 0 0 0 0 <= Grid 0 1 0
If we fix the path length to 4, we get 16 possible paths from A to B
Adding up every visited square (divided by the total number of paths) gives us the probability of visiting a square while walking along a random path
0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
0 1 1 2 1 0 1 1 2 0 0 1 2 2 0 0 1 1 2 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0
+ + + +
0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
0 2 1 1 0 0 2 1 1 0 1 2 1 1 0 0 2 2 1 0
0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
+ + + +
0 1 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 0
0 1 0 1 0 0 1 1 1 0 0 1 1 1 0 0 1 2 1 0
0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0
+ + + +
0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 3 4 3 0
0 1 0 1 0 + 0 1 1 1 0 + 0 1 1 1 0 + 0 1 2 1 0 = 1 20 18 20 1
0 1 1 1 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 3 4 3 0
There are a number of ways to tackle the problem on a grid
- purely combinatorial (unwieldy formulas with many binomials, not very insightful)
- via adjacency matrix, which to the n-th power gives the reachable positions after n steps (very powerful, one can easily implement obstacles on the grid)
- convolving a zero-filled grid with a
1on starting position with the movement kernel n times (basiclly the same but less efficient, though easier to implement and visualize)
How does it extend for a continuos path on a plane?
My idea was to approximate "movement in any direction by $\Delta r$" with the same grid approach and a discretized ring as movement kernel
Keeping $\Delta r = 7$ (second ring) constant while increasing the path length and distance between the two points is equivalent to making $\Delta r$ smaller without amplifying the discretization artifacts. (shown is log(1 + #pixel_visits) because the numbers are huge, in the last image there is a total of $10^{198}$ possible paths)
As one would expect the outer border is elliptical. The insides look nice and smooth, somewhat like a sum of ellipses with same focal points and different excentrities, giving hope that an analytical solution might exist, but how to find it?
The question needs some additional information:
1 ) The result will be different in 2D or 3D. So let´s assume 2D.
2 ) The results will depend on the metric you use to differentiate paths. In your example, you use a uniform grid to discretize the problem. This "solves" the problem, but then it is not really continuous (only asymptotically when the pixel size tends to zero). Let´s assume that the points are placed in a uniformly distributed 2D Cartesian grid.
3 ) The result will also depend on the metric you use to define the length. For instance, with the adjacency matrix, if only pixels connected in the horizontal and vertical directions (i.e. not diagonally) are defined to be adjacents, then the result will correspond to the Manhattan Metric (not Euclidean). So let´s assume that the lengths of the paths are defined by the Euclidean metric.
With all this defined, the solution you proposed based on convolutions is interesting, but I think it needs a minor but significant correction:
In order to avoid discretization errors, you use a ring with r=7 pixels, considering that the location of the points and lenghts can be dilated by same factor. Nevertheless, if the two reference points are located in a single pixel before the dilation, they have to be dilated as well, and they will no longer be one pixel but a circle. Doing this will correct the missing pixels of your first image. By using your ring with radius=7 pixels, you are assuming that your pixels are smaller than what you can actually observe (this is related to my point #2), and therefore, all paths that starts or end within the ring are "the same".
As you mentioned, you can reduce the radius of the ring relative to the distance between the points by increasing this distance a factor so large that the errors from the discretization becomes negligible, but I think that in any case, the points should be dilated to keep the consistency of your argument.