During this year's Distributed Code Jam I had an idea for solving the hard part of Toothpick Sculptures, but had no time to implement it, or even to work out all the details. When the official solution was presented, I was surprised that it was very different to mine. Recently Pablo from Google has been able to help me check that my solution does indeed pass the system tests (thanks Pablo), so it seems like a good time to describe the solution.

There are going to be a lot of different trees involved, so let's give them names. A "minitree" is one of the original sculptures of N toothpicks, and its root is a "miniroot". The "maxitree" is the full tree of 1000N toothpicks. Let's assume that each minitree is built out of toothpicks of one colour. We can then consider a tree whose vertices are colours, called the "colour tree", which contains an edge between two colours if and only if the maxitree contains an edge between two vertices of those colours.

For each colour, we can find its children in the colour tree by walking the corresponding minitree looking for children which are miniroots. This can be done in parallel over the colours.

For a given vertex, we can use DP to solve the following two problems for the subtree: what is the minimum cost to stabilise the subtree if we stabilise the root, and what is the minimum cost if we do not stabilise the root? But this will not be so easy to parallelise. We can go one step further: if we fix some vertex C and cut off the subtree rooted at C, then we can answer the same two problems for each case of C being stabilised or not stabilised (four questions in total). If we later obtain the answers to the original two queries for C, then we can combine this with the four answer we have to answer the original queries for the full tree.

This is sufficient to solve the small problem, and more generally any test case where the colour tree does not branch. For each miniroot, compute the DP, where the miniroot corresponding to the single child (if any) in the colour tree is the cutoff. These DPs can be done in parallel, and the master can combine the results.

Let's consider another special case, where the colour tree is shallow. In this case, one can solve one layer at a time, bottom up, without needing to use the cutoff trick at all. The colours in each layer of the colourtree are independent and can be solved in parallel, so the latency is proportional to the depth of the tree. The results of each layer are fed into the calculations of the layer above.

So, we have a way to handle long paths, and a way to handle shallow trees. This should immediately suggest light-heavy decomposition. Let the "light-depth" of a node in the colour tree be the number of light edges between it and the root. The native of light-heavy decomposition guarantees that light-depth is at most logarithmic in the tree size (which is 1000). We will process all nodes with the same light-depth in parallel. This means that a node in a tree may be processed at the same time as its children, but only along a heavy path. We thus handle each heavy path using the same technique as in the small problem. For other children in the colour tree, the subtree results were computed in a previous pass and are sent to the slave.

There are going to be a lot of different trees involved, so let's give them names. A "minitree" is one of the original sculptures of N toothpicks, and its root is a "miniroot". The "maxitree" is the full tree of 1000N toothpicks. Let's assume that each minitree is built out of toothpicks of one colour. We can then consider a tree whose vertices are colours, called the "colour tree", which contains an edge between two colours if and only if the maxitree contains an edge between two vertices of those colours.

For each colour, we can find its children in the colour tree by walking the corresponding minitree looking for children which are miniroots. This can be done in parallel over the colours.

For a given vertex, we can use DP to solve the following two problems for the subtree: what is the minimum cost to stabilise the subtree if we stabilise the root, and what is the minimum cost if we do not stabilise the root? But this will not be so easy to parallelise. We can go one step further: if we fix some vertex C and cut off the subtree rooted at C, then we can answer the same two problems for each case of C being stabilised or not stabilised (four questions in total). If we later obtain the answers to the original two queries for C, then we can combine this with the four answer we have to answer the original queries for the full tree.

This is sufficient to solve the small problem, and more generally any test case where the colour tree does not branch. For each miniroot, compute the DP, where the miniroot corresponding to the single child (if any) in the colour tree is the cutoff. These DPs can be done in parallel, and the master can combine the results.

Let's consider another special case, where the colour tree is shallow. In this case, one can solve one layer at a time, bottom up, without needing to use the cutoff trick at all. The colours in each layer of the colourtree are independent and can be solved in parallel, so the latency is proportional to the depth of the tree. The results of each layer are fed into the calculations of the layer above.

So, we have a way to handle long paths, and a way to handle shallow trees. This should immediately suggest light-heavy decomposition. Let the "light-depth" of a node in the colour tree be the number of light edges between it and the root. The native of light-heavy decomposition guarantees that light-depth is at most logarithmic in the tree size (which is 1000). We will process all nodes with the same light-depth in parallel. This means that a node in a tree may be processed at the same time as its children, but only along a heavy path. We thus handle each heavy path using the same technique as in the small problem. For other children in the colour tree, the subtree results were computed in a previous pass and are sent to the slave.