Facebook Hacker Cup 2015 has completed its weekly routine with Round 3 on Saturday night (problems with Facebook login, results with Facebook login, top 5 on the left, my screencast). Top 25 will travel to the United States for the final round on March 6, and big congratulations to Gleb who claimed the first place in this really competitive round!

Let's also come back to the interesting problem I mentioned in last week's summary: you are given a directed graph where each arc has a cost, one of the vertices is marked as starting point, and one as ending point. You need to pick a subset of arcs with the smallest total cost such that every path from the starting point to the ending point (including paths that pass through the same vertex or arc more than once) passes through arcs from this subset exactly once.

Whenever we need to pick a subset of arcs that splits a graph in two, the theory of flows and cuts comes to mind. However, we're not simply asked to find a minimum cut on a given graph, and not even on its condensation, as illustrated by the following acyclic example:

In this graph, the minimum cut is of size 2 and separates vertices S and A from vertices B and T. However, the path S-B-A-T crosses the cut three times (twice going outside, and once going inside), and we need for each path to cross the cut exactly once. Basically, the minimum cut is solving the "at least once" version.

But the power of the flow/cut theory often lies in applying it to slightly modified graphs. For cuts in particular, adding infinite arcs to the graph is usually the solution of choice. Here our goal is to make sure one can't re-enter the cut after leaving it, so for each arc we'll add a reverse arc with infinite capacity, as illustrated by the following picture:

In particular, the newly added A-B arc of infinite capacity changes the incorrect cut displayed above from capacity 2 to infinite capacity (1+1+infinity), and thus the smallest cut is now of size 3, displayed by the dotted line to the left. There's also another cut of size 3 - just cut off the source vertex S. It's not hard to verify that every path from S to T croses this cut exactly once.

Another great thing about this addition of infinite arcs is that it allows us to skip the condensation step. If there's a cycle somewhere in our graph, we'll thus add a reverse infinite cycle, and thus no minimum cut will ever cross any cycle!

Such observations make this solution "click" when you see it: instead of fighting with the problem, everything flows naturally and the solution makes sense from all angles. I especially value such moments, I think they show the beauty of mathematics and algorithms.

However, everything was not that simple in this problem :) It turns out that in order for this solution to work correctly in all cases, we need to prune the graph before adding the infinite arcs. More specifically, we need to remove all vertices that don't lie on any path from S to T from the graph. Can you see where we went wrong, and why doesn't the cut theory save us in this case?

Thanks for reading, and check back next week!

Let's also come back to the interesting problem I mentioned in last week's summary: you are given a directed graph where each arc has a cost, one of the vertices is marked as starting point, and one as ending point. You need to pick a subset of arcs with the smallest total cost such that every path from the starting point to the ending point (including paths that pass through the same vertex or arc more than once) passes through arcs from this subset exactly once.

Whenever we need to pick a subset of arcs that splits a graph in two, the theory of flows and cuts comes to mind. However, we're not simply asked to find a minimum cut on a given graph, and not even on its condensation, as illustrated by the following acyclic example:

In this graph, the minimum cut is of size 2 and separates vertices S and A from vertices B and T. However, the path S-B-A-T crosses the cut three times (twice going outside, and once going inside), and we need for each path to cross the cut exactly once. Basically, the minimum cut is solving the "at least once" version.

But the power of the flow/cut theory often lies in applying it to slightly modified graphs. For cuts in particular, adding infinite arcs to the graph is usually the solution of choice. Here our goal is to make sure one can't re-enter the cut after leaving it, so for each arc we'll add a reverse arc with infinite capacity, as illustrated by the following picture:

In particular, the newly added A-B arc of infinite capacity changes the incorrect cut displayed above from capacity 2 to infinite capacity (1+1+infinity), and thus the smallest cut is now of size 3, displayed by the dotted line to the left. There's also another cut of size 3 - just cut off the source vertex S. It's not hard to verify that every path from S to T croses this cut exactly once.

Another great thing about this addition of infinite arcs is that it allows us to skip the condensation step. If there's a cycle somewhere in our graph, we'll thus add a reverse infinite cycle, and thus no minimum cut will ever cross any cycle!

Such observations make this solution "click" when you see it: instead of fighting with the problem, everything flows naturally and the solution makes sense from all angles. I especially value such moments, I think they show the beauty of mathematics and algorithms.

However, everything was not that simple in this problem :) It turns out that in order for this solution to work correctly in all cases, we need to prune the graph before adding the infinite arcs. More specifically, we need to remove all vertices that don't lie on any path from S to T from the graph. Can you see where we went wrong, and why doesn't the cut theory save us in this case?

Thanks for reading, and check back next week!

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