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How to convert an unbalanced tree into a (balanced) spanning tree? Suppose I have a tree (with different (not necessarily distinct) number of children at different nodes). I want to manipulate the tree in such a way that it becomes a k-ary spanning tree.

Various iterations on the tree are allowed. The restriction is that we cannot just collect all the nodes at one place and then make a spanning tree out of them (which would be a trivial way to do). Rather the spanning tree has to be created from the given tree. That is the children can exchange information (e.g. the number of child nodes it has and the ids of the child nodes) with the parent (and the grandparent, if required) and the parent makes decision to move the nodes between its children (in order to balance the tree).

You may have understood that I am trying to do this in a parallel computing environment. Where, all a node knows is the id of its parent, its children and the number of nodes in each subtree with its children as the root.

(Parent and children will change as we try to balance the tree). Any hint on how to approach this problem?

Reply to the comment that why is this problem important/ worth considering - after all the trivial apporach is scalable:

  1. Theoretically it is challenging to develop an algorithm that uses lesser than O(N) space(used in the trivial approach) to build the spanning tree.

  2. It is interesting to think about alternative solution approaches at scale.

  3. As far as numbers are concerned: N=100,000 (which is common in today's supercomputers, N will be 1000,000 in the upcoming BG/Q). In the trivial approach steps invloved are a) all-reduce b) O(N) to construct the spanning tree and c) and finally a one-to-many broadcast.

An alternative distributed approach may not give much improvement but out of curosity it may be is worth trying.

Josh Lee
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Akhil
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  • Why we do not balance the tree like AVL tress? – Pih May 15 '11 at 22:29
  • AVL trees are balanced binary search trees. In this problem we just have a random tree with a node having any number of children (which is not the case in AVL trees). The resulting tree does not have to follow any other property other than being a balanced k-ary spanning tree. – Akhil May 16 '11 at 01:47
  • Why do you think there's a k-ary spanning tree possible? Imagine a graph which is just a linear chain. The spanning tree is then an isomorphic copy of the linear chain (with arbitrarily chosen root node). You can't make a K-ary tree from it. You have at least soften your requirements to produce a tree with minimum depth. – Ira Baxter May 16 '11 at 01:57
  • @Ira: I think you've misunderstood a little bit. It seems that the connectivity of the nodes is allowed to change. I work in the same field as the asker (apparently), and I see no reason it wouldn't be. That change in connectivity just needs to be communicated among the affected nodes. – Phil Miller May 16 '11 at 02:09
  • @Novelocrat: The connectivity of the original graph can change? If this is the situation, I don't understand what it means (or even the utility) to have a spanning tree if you can change the underlying graph arbitrarily. If you mean the arcs of the spanning tree can change, I'm not surprised; I thought that was the point of the question. In that case, my objection still stands. – Ira Baxter May 16 '11 at 03:04
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    @Ira: Yes, the connectivity of the graph can change by leaps and bounds. The underlying graph represents nothing but an unbalanced connected tree, we can add new edges to make it a balanced k-ary tree. The challenge in the question is to create the spanning tree without collecting all the node-ids at a single place – Akhil May 16 '11 at 17:32
  • Per your 'answer', I'm closing this. – Tim Post Oct 15 '11 at 02:13

2 Answers2

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Respectfully, I think you're seriously underestimating the extent to which your "trivial case" scales in a typical parallel computing environment. Care to post some actual numbers?

sped
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  • I Agree that power of the trivial case has not been given the due credit. Still, I have added the reply as a part of the question. These are the only reasons that I could think of. Do you have any points in favor of the problem ? – Akhil May 16 '11 at 18:30
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Some random musings at the pieces of a suitable algorithm:

  1. Select a 'root' arbitrarily, perhaps as the lowest-index element among those participating, or perhaps as the root of the existing structure.
  2. Think of a 'phase' as one parallel reduction that computes the depth and identity of the deepest part of the tree and the shallowest part of the tree that isn't already saturated.
  3. After each phase, the root directs the movement of the right number of nodes from the deepest part to the shallow part to balance by depth.
  4. Repeat until fully balanced

This might also work in a fully distributed fashion, where each neighborhood balances itself according to something like the AVL criterion, and eventually large structural changes propagate through.

Phil Miller
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