Undirected graph conversion to tree

Question

Given an undirected graph in which each node has a Cartesian coordinate in space that has the general shape of a tree, is there an algorithm to convert the graph into a tree, and find the appropriate root node?

Note that our definition of a "tree" requires that branches do not diverge from parent nodes at acute angles.

See the example graphs below. How do we find the red node?

Answer 1

here is a suggestion on how to solve your problem.

prerequisites

• notation:
  
  • g graph, g.v graph vertices
  • v,w,z: individual vertices
  • e: individual edge
  • n: number of vertices
• any combination of an undirected tree g and a given node g.v uniquely determines a directed tree with root g.v (provable by induction)

idea

• complement the edges of g by orientations in the directed tree implied by g and the yet-to-be-found root node by local computations at the nodes of g.
• these orientations will represent child-parent-relationsships between nodes (v -> w: v child, w parent).
• the completely marked tree will contain a sole node with outdegree 0, which is the desired root node. you might end up with 0 or more than one root node.

algorithm

assumes standard representation of the graph/tree structure (eg adjacency list)

1. all vertices in g.v are marked initially as not visited, not finished.

2. visit all vertices in arbitrary sequence. skip nodes marked as 'finished'.
   let v be the currently visited vertex.

   • 2.1 sweep through all edges linking v clockwise starting with a randomly chosen e_0 in the order of the edges' angle with e_0.

   • 2.2. orient adjacent edges e_1=(v,w_1), e_2(v,w_2), that enclose an acute angle.
     adjacent: wrt being ordered according to the angle they enclose with e_0.

     [ note: the existence of such a pair is not guaranteed, see 2nd comment and last remark. if no angle is acute, proceed at 2. with next node. ]

     • 2.2.1 the orientations of edges e_1, e_2 are known:

       • w_1 -> v -> w_2: impossible, as a grandparent-child-segment would enclose an acute angle
       • w_1 <- v <- w_2: impossible, same reason

       • w_1 <- v -> w_2: impossible, there are no nodes with outdegree >1 in a tree

       • w_1 -> v <- w_2:
         only possible pair of orientations. e_1, e_2 might have been oriented before. if the previous orientation violates the current assignment, the problem instance has no solution.

     • 2.2.2 this assignment implies a tree structure on the subgraphs induced by all vertices reachable from w_1 (w_2) on a path not comprising e_1 (e_2`). mark all vertices in both induced subtrees as finished

       [ note: the subtree structure might violate the angle constraints. in this case the problem has no solution. ]

   • 2.3 mark v visited. after completing steps 2.2 at vertex v, check the number nc of edges connecting that have not yet been assigned an orientation.

     • nc = 0: this is the root you've been searching for - but you must check whether the solution is compatible with your constraints.

     • nc = 1: let this edge be (v,z).
       the orientation of this edge is v->z as you are in a tree. mark v as finished.

       • 2.3.1 check z whether it is marked finished. if it is not, check the number nc2 of unoriented edges connecting z. nc2 = 1: repeat step 2.3 by taking z for v.

3. if you have not yet found a root node, your problem instance is ambiguous: orient the remaining unoriented edges at will.

remarks

1. termination: each node is visited at max 4 times:

   • once per step 2
   • at max twice per step 2.2.2
   • at max once per step 2.3

2. correctness:

   • all edges enclosing an acute angle are oriented per step 2.2.1

3. complexity (time):

   • visiting every node: O(n);

   • the clockwise sweep through all edges connecting a given vertex requires these edges to be sorted.
     thus you need O( sum_i=1..m ( k_i * lg k_i ) ) at m <= n vertices under the constraint sum_i=1..m k_i = n.

     in total this requires O ( n * lg n), as sum_i=1..m ( k_i * lg k_i ) <= n * lg n given sum_i=1..m k_i = n for any m <= n (provable by applying lagrange optimization).

     [ note: if your trees have a degree bounded by a constant, you theoretically sort in constant time at each node affected; grand total in this case: O(n) ]

   • subtree marking:
     each node in the graph is visited at max 2 times by this procedure if implemented as a dfs. thus a grand total of O(n) for the invocation of this subroutine.

     in total: O(n * lg n)

4. complexity (space):

   • O(n) for sorting (with vertex-degree not constant-bound).

5. problem is probably ill-defined:

   • multiple solutions: e.g. steiner tree
   • no solution: e.g. graph shaped like a double-tipped arrow (<->)

Answer 2

A simple solution would be to define a 2d rectangle around the red node or the center of your node and compute each node with a moore curve. A moore curve is a space-filling curve, more over a special version of a hilbert curve where the start and end vertex is the same and the coordinate is in the middle of the 2d rectangle. In generell your problem looks like a discrete addressing space problem.