As I pointed out to @Voss and @Star Strider, if the points in B mst be unique, then there is an issue. If you simply choose the closest points (where it might be best to use knnsearch anyway) then one of the points in B may be simultaneously closest to different points in C.
The simple answer is to test to see if there are any replicate points chosen, where point k from B is simultaneously closest to two (or more!) points in C. But then you need to decide which other point to add to the set. I would probably do this by a resampling scheme, where all of the points already chosen are effectively removed from B. Now find a new point, which is the next closest to the point from C which was impacted. The code would not be too complex, and it would be reasonably efficient.
I'll give an example. Here, I'll use a smaller set of points to make thigns easier to visualize, and also to make it more likely there will be a conflict, I have made the size of B smaller. Larger sets will of course reduce the probability of a failure.
C = rand(10,2);
B = rand(15,2);
[idxB,D] = knnsearch(B,C,K=1)
In this set, you can see that point # 1 from B was simultaneously closest to 4 different points in C. And point 13 was also duplicated in that list. We could test to see if there were any replicates simply enough. For example,
if numel(unique(idxB)) < size(C,1)
% stuff
end
But now there is an ambiguity. Point #1 from B was closest to point 7 in C. Do we now choose a second best point from B for points [2 3 8] in C? What if there is again an issue? Is there some optimal set which is "best" in some unspecified way? And this is where your problem specification fails, because it does not suggest how these issues must be addressed.
Will this be a frequent problem? SURE AS HELL YES! Consider this simulation.
failcount = 0;
for i = 1:100
C = rand(60,2);
B = rand(600,2);
[idxB,D] = knnsearch(B,C,k=1);
if numel(unique(idxB)) ~= size(C,1)
failcount = failcount + 1;
end
end
failcount
failcount =
99
So out of 100 random sets generated, 99 of them had a conflict. If uniqueness is an issue, then you WILL need to write code that worries about these conflicts.
Anyway, COULD you resolve all conflicts using a loop? Well, yes.
C = rand(60,2);
B = rand(600,2);
idxB = NaN(size(C,1),1);
Btemp = B;
for i = 1:size(C,1)
idxB(i) = knnsearch(Btemp,C(i,:),k=1);
Btemp(idxB(i),:) = inf;
end
numel(unique(idxB))
ans =
60
And clearly this will result in a unique set, that has no conflicts.
