## Sunday, November 20, 2016

### Topcoder Open 2016 Semifinal 1

Here are my solutions to the first semifinal of the TCO 2016 algorithm contest.

### Easy

Let's start with some easy cases. If s = t, then the answer is obviously zero. Otherwise, it is at least 1. If s and t share a common factor, then the answer is 1, otherwise it is at least 2.
We can eliminate some cases by noting that some vertices are isolated. Specifically, if either s or t is 1 or is a prime greater than n/2 (and s≠t) then there is no solution.
Assuming none of the cases above apply, can the answer be exactly 2? This would mean a path s → x → t, where x must have a common factor with s and a (different) common factor with t. The smallest x can be is thus the product of the smallest prime factors of s and t. If this is at most n, then we have a solution of length 2, otherwise the answer is at least 3.
If none of the above cases applies, then answer is in fact 3. Let p be the smallest prime factor of s and q be the smallest prime factor of t. Then the path s → 2p → 2q → t works (note that 2p and 2q are at most n because we eliminated cases where s or t is a prime greater than n/2 above).

### Medium

We can create a graph with the desired properties recursively. If n is 1, then we need just two vertices, source and sink, with an edge from source to sink of arbitrary weight.
If n is even, we can start with a graph with n/2 min cuts and augment it. We add a new vertex X and edges source → X and X → sink, both of weight 1. For every min cut of the original graph, X can be placed on either side of the cut to make a min cut for the new graph.
If n is odd, we can start with a graph with n - 1 min cuts and augment it. Let the cost of the min cut of the old graph be C. We create a new source, and connect it to the original source with an edge of weight C. The cost of the min cut of this new graph is again C (this can easily be seen by considering the effect on the max flow rather than the min cut). There are two ways to achieve a cut of cost C: either cut the new edge only, or make a min cut on the original graph.
We just need to check that this will not produce too many vertices. In the binary representation of n, we need a vertex per bit and a vertex per 1 bit. n can have at most 10 bits and at most 9 one bits, so we need at most 19 vertices.

### Hard

I made a complete mess of this during the contest, solving the unweighted version and then trying to retrofit it to handle a weighted version. That doesn't work.

Here's how to do it, which I worked out based on some hints from tourist and coded at 3am while failing to sleep due to jetlag. Firstly, we'll make a change that makes some of the later steps a little easier: for each pair of adjacent vertices, add an edge with weight ∞. This won't violate the nesting property, and guarantees that there is always a solution. Let the length of an edge be the difference in indices of the vertices (as opposed to the cost, which is an input). Call an edge right-maximal if it's the longest edge emanating from a particular vertex, and left-maximal if it is the longest edge terminating at a particular vertex. A useful property is that an edge is always either right-maximal, left-maximal, or both. To see this, imagine an edge that is neither: it must then be nested inside longer edges terminating at each end, but these longer edges would cross, violating the given properties.

We can now make the following observations:
1. If x → y is right-maximal and color[x] is used, then color[y] must be used too. This is because the path must pass through x, and after x cannot skip over y.
2. If x → y is the only edge leaving x, then the edge will appear in a path if and only if color[x] is used.
3. If x → y is right-maximal and x →  z is the next-longest edge from x, then x → y is in a path if and only if color[x] is used and color[z] is not used.
These statements all apply similarly for left-maximal edges of course. We can consider vertices 0 and n to be of colour 0 in the above, which is a color that must be picked. What is less obvious is that condition 1 (together with its mirror) is sufficient to determine whether a set of colours is admissible. Suppose we have a set of colours that satisfies condition 1, and consider the subset S of vertices whose colours come from this set. We want to be sure that every pair of consecutive vertices in S are linked by an edge. Suppose there is a pair (u, v) which are not. Consider the right-maximal edge from u: by condition 1, it must terminate at some vertex in S, which must then be to the right of v. But the left-maximal edge to v must start to the left of u, and so these two edges cross.

We can now use this information to construct a new graph, whose minimum cut will give us the cost of the shortest path. It has a vertex per colour, a source vertex corresponding to the special colour 0, and a sink vertex not corresponding to a colour. The vertices on the source side of the cut correspond to the colours that are visited. Every time colour C implies colour D, add an edge C → D of weight ∞, to prevent C being picked while D is not. If an edge falls into case 2 above, add an edge from color[x] to the sink with weight equal to the edge cost. Otherwise it falls into case 3; add an edge from color[x] to color[y] with weight equal to the edge cost. Each edge in the original graph now corresponds to an edge in this new graph, and the original edge is traversed if and only if the new edge forms part of the cut.