How to compute the visible area based on a heightmap? - graphics

Question

I have a heightmap. I want to efficiently compute which tiles in it are visible from an eye at any given location and height.
This paper suggests that heightmaps outperform turning the terrain into some kind of mesh, but they sample the grid using Bresenhams.
If I were to adopt that, I'd have to do a line-of-sight Bresenham's line for each and every tile on the map. It occurs to me that it ought to be possible to reuse most of the calculations and compute the heightmap in a single pass if you fill outwards away from the eye - a scanline fill kind of approach perhaps?
But the logic escapes me. What would the logic be?
Here is a heightmap with a the visibility from a particular vantagepoint (green cube) ("viewshed" as in "watershed"?) painted over it:
Here is the O(n) sweep that I came up with; I seems the same as that given in the paper in the answer below How to compute the visible area based on a heightmap? Franklin and Ray's method, only in this case I am walking from eye outwards instead of walking the perimeter doing a bresenhams towards the centre; to my mind, my approach would have much better caching behaviour - i.e. be faster - and use less memory since it doesn't have to track the vector for each tile, only remember a scanline's worth:
typedef std::vector<float> visbuf_t;
inline void map::_visibility_scan(const visbuf_t& in,visbuf_t& out,const vec_t& eye,int start_x,int stop_x,int y,int prev_y) {
const int xdir = (start_x < stop_x)? 1: -1;
for(int x=start_x; x!=stop_x; x+=xdir) {
const int x_diff = abs(eye.x-x), y_diff = abs(eye.z-y);
const bool horiz = (x_diff >= y_diff);
const int x_step = horiz? 1: x_diff/y_diff;
const int in_x = x-x_step*xdir; // where in the in buffer would we get the inner value?
const float outer_d = vec2_t(x,y).distance(vec2_t(eye.x,eye.z));
const float inner_d = vec2_t(in_x,horiz? y: prev_y).distance(vec2_t(eye.x,eye.z));
const float inner = (horiz? out: in).at(in_x)*(outer_d/inner_d); // get the inner value, scaling by distance
const float outer = height_at(x,y)-eye.y; // height we are at right now in the map, eye-relative
if(inner <= outer) {
out.at(x) = outer;
vis.at(y*width+x) = VISIBLE;
} else {
out.at(x) = inner;
vis.at(y*width+x) = NOT_VISIBLE;
}
}
}
void map::visibility_add(const vec_t& eye) {
const float BASE = -10000; // represents a downward vector that would always be visible
visbuf_t scan_0, scan_out, scan_in;
scan_0.resize(width);
vis[eye.z*width+eye.x-1] = vis[eye.z*width+eye.x] = vis[eye.z*width+eye.x+1] = VISIBLE;
scan_0.at(eye.x) = BASE;
scan_0.at(eye.x-1) = BASE;
scan_0.at(eye.x+1) = BASE;
_visibility_scan(scan_0,scan_0,eye,eye.x+2,width,eye.z,eye.z);
_visibility_scan(scan_0,scan_0,eye,eye.x-2,-1,eye.z,eye.z);
scan_out = scan_0;
for(int y=eye.z+1; y<height; y++) {
scan_in = scan_out;
_visibility_scan(scan_in,scan_out,eye,eye.x,-1,y,y-1);
_visibility_scan(scan_in,scan_out,eye,eye.x,width,y,y-1);
}
scan_out = scan_0;
for(int y=eye.z-1; y>=0; y--) {
scan_in = scan_out;
_visibility_scan(scan_in,scan_out,eye,eye.x,-1,y,y+1);
_visibility_scan(scan_in,scan_out,eye,eye.x,width,y,y+1);
}
}
Is it a valid approach?
it is using centre-points rather than looking at the slope between the 'inner' pixel and its neighbour on the side that the LoS passes
could the trig in to scale the vectors and such be replaced by factor multiplication?
it could use an array of bytes since the heights are themselves bytes
its not a radial sweep, its doing a whole scanline at a time but away from the point; it only uses only a couple of scanlines-worth of additional memory which is neat
if it works, you could imagine that you could distribute it nicely using a radial sweep of blocks; you have to compute the centre-most tile first, but then you can distribute all immediately adjacent tiles from that (they just need to be given the edge-most intermediate values) and then in turn more and more parallelism.
So how to most efficiently calculate this viewshed?

Answer 1

What you want is called a sweep algorithm. Basically you cast rays (Bresenham's) to each of the perimeter cells, but keep track of the horizon as you go and mark any cells you pass on the way as being visible or invisible (and update the ray's horizon if visible). This gets you down from the O(n^3) of the naive approach (testing each cell of an nxn DEM individually) to O(n^2).
More detailed description of the algorithm in section 5.1 of this paper (which you might also find interesting for other reasons if you aspire to work with really enormous heightmaps).