Question

let there be an undirected tree T, and let there be: T.leaves – all the leaves (each v such that d(v) = 1). we know: |T.leaves| and the distance between u and v for each u,v in T.leaves.

in other words: we have an undirected tree, and we know how many leaves it has, and the distance between every 2 leaves.

we need to find how many inside vertices (d(v)>1) are in the tree.

note: building the complete tree is impossible because if we have only 2 leaves but the distance between them is 2^30, it will take too long…

I tried to start from the shortest distance and count how many vertices are between them, and then adding the vertex closest to them, but for this I need some formula f(leaves_counted,next_leaf) but I could not manage to find that f…

any ideas?

Answer 1

Continued from discussion in comments. This is how to check a particular (compressed) edge to see if you can attach the new vertex n somewhere in the middle of it, without iterating over the distances.

Ok, so you need to find three numbers: l (the distance of the attach point from the left node of the edge in question), x (the distance of the new node from the attach point) and r (symmetrical to l.)

Obviously, for every node y in set L (the left part of the tree), its distance to A must differ from its distance to n by the same number (lets call it dl which must be equal l + x). If this is not the case, there is no solution for this particular edge. Same goes for nodes in R, with dr and r + x respectively.

If the above holds, then you have three equations:

l + x = dl

r + x = dr

r+l = dist(A,B)

Three equations, three numbers. If this has a solution then you have found the right edge.

At worst you need to iterate the above for every edge, but I think it can be optimized – the distance check on L and R might exclude one of the parts of the tree from further search. It might also be possible to somehow get the number of nodes without even constructing the tree.