A few days ago, a redditor asked the people of /r/TheWire to rank the five seasons of the HBO television show “The Wire”, and several people posted their rankings. I wanted to take their individual rankings and uncover one aggregate ranking that captured the overall preferences of the entire group of posters. In other words, I wanted to implement a preferential voting scheme.
I was aware of Arrow’s theorem, which says that no voting scheme can satisfy a certain small set of reasonable criteria when voters are ranking three or more alternatives. However, I was also aware that a number of countries employ preferential voting systems. So, I didn’t give up, and I searched for a preferential voting system that would fit the bill.
I found one that worked: the ranked pairs voting scheme. In the ranked pairs voting system, you first count all the times that voters ranked Alternative X ahead of Alternative Y, for all pairs of alternatives (X, Y). If voters ranked Alternative X ahead of Alternative Y more often, then Alternative X is a majority over Alternative Y.
Next you sort the majorities. For example, say that voters ranked Alternative X ahead of Alternative Y three times and that they ranked Alternative A ahead of Alternative B five times. Then the latter majority comes before the former majority in the sort.
But what if the list of majorities creates a cycle? For example, say that we have [(A, B), (B, C), (C, A)]. Then A is a majority over B and B is a majority over C. But C is a majority over A. So, by transitivity, A is a majority over A. But an alternative X can be a majority over an alternative Y only if X was ranked ahead of Y, and A can’t be ranked ahead of itself.
So, here’s how you resolve that issue. You treat the list of majorities like a directed acyclic graph where the dag includes a directed edge from X to Y if X is a majority over Y. And you just go down the sorted list of majorities. For each majority, you try to add a corresponding edge to the dag. If adding that edge would produce a cycle, then you don’t add it. If it wouldn’t produce a cycle, then you do add it, and that majority gets locked in to the final list of majorities.
The winner of the election is the source of the dag, i.e., the vertex that doesn’t have any incoming edges. But I didn’t want to know just the winner. I wanted to know the aggregate ranking. So, I sorted the alternatives by the number of outgoing edges that their corresponding vertices have in the dag. The number of outgoing edges of a vertex corresponds to the number of alternatives that an alternative is a majority over.
I collected the data on the night of September 1. I simply scrolled down the list of comments and recorded the rankings that commenters reported. The only complication was that a couple of commenters expressed indifference between some seasons. So, for example, one person wrote, “4 > 1 > 3/2 > 5”. In such a case, I ranked seasons in the order that they were written. So, in this particular case, I placed Season 3 before Season 2.
Then I ran the program, and the result was 4 > 1 > 2 > 3. Season 5 didn’t even make the list.
Earlier today I searched through the ACM-ICPC archive of problems that have been used in ACM-ICPC Regionals and World Finals, looking for some interesting problems to solve. I found one that reminded me of another ACM-ICPC problem that I solved. You can read the full description of it, but I’ll put it in a nutshell:
You’re given a stack of playing cards. You can swap adjacent cards. You want to find the minimum number of swaps it takes to sort the cards so that all the diamonds and hearts in the stack appear before all the spades and clubs.
I realized that you are searching for a certain state of the stack of cards. Because you want to find that state in the minimum number of swaps, you can perform a version of breadth-first search to solve the problem.
I remembered the Marbles in Three Baskets problem, another ACM-ICPC problem that I solved. I applied the same basic technique and philosophy of problem-solving to this problem that I outline in the post about the Marbles in Three Baskets problem.
In the vertex coloring problem, you are given a graph and a set of colors. You want to assign colors to the vertices of the graph in such a way that no adjacent vertices have the same color.
The vertex coloring problem appears in a lot of different forms in the real world. To take a rather frivolous example, just consider Sudoku. Each cell corresponds to a vertex, the numbers 1-9 correspond to colors that you can assign to the vertices, and two vertices are adjacent if the corresponding cells are in the same row, column, or subgrid.
The vertex coloring problem is a constraint satisfaction problem. In a constraint satisfaction problem, you are given a set of variables, a set of possible values for each variable, and a set of constraints. You want to find an assignment of values to the variables that satisfies the constraints.
In the vertex coloring problem, the vertices are the variables, the colors are the possible values for each variable, and the constraint is that no adjacent vertices can have the same color.
There are a number of ways to solve constraint satisfaction problems. One way is recursive backtracking. The n queens problem is another constraint satisfaction problem, and I have applied recursive backtracking to solve the n queens problem.
Today I want to talk about the min-conflicts algorithm. Like simulated annealing, which I have also talked about, the min-conflicts algorithm is a type of local search. You start with some assignment of values to variables. Unless you get really lucky, that initial assignment will produce conflicts. For example, if a vertex is colored green and two of its neighboring vertices are colored green, then its being colored green produces two conflicts. So, at each iteration, the min-conflicts algorithm picks a conflicting variable at random and assigns to it a value that produces the minimal number of conflicts. If there is more than one such value, then it chooses one at random. It then repeats this process again and again, until it either finds a complete assignment of values that satisfies all of the constraints or goes through the maximum number of iterations that it may go through.
I implemented the min-conflicts algorithm, and I applied it to the vertex covering problem. I also decided to go back and apply the min-conflicts algorithm to the n queens problem. You can find more of my programs at my GitHub.
This week /r/DailyProgrammer posed a series of challenges related to graph theory.
The first challenge gives you the number of vertices of an undirected graph and the graph’s edges. It asks you to find the degree of each vertex. The bonus challenge asks you to construct the adjacency matrix of the graph. I solved both problems, and I combined my solutions into one solution.
Each vertex i corresponds to a row in the adjacency matrix. For each column j in the matrix, if there is an edge between vertex i and vertex j, then the matrix entry i,j is equal to 1, and 0 otherwise. So, to find the number of edges for a vertex in an undirected graph, you can just construct the adjacency matrix and then sum all of the entries of the corresponding row.
The eccentricity of a vertex is the greatest distance from that vertex to any other vertex. So, to find the radius and the diameter of a graph, you can find the eccentricity of each vertex. Then the radius is the smallest eccentricity, and the diameter is the largest eccentricity.
Because I find the eccentricity of each vertex, I find the distances between all pairs of vertices in the graph. So, my first thought was to apply the Floyd-Warshall algorithm. However, one of the given graphs has a cycle, and I mistakenly believed that Floyd-Warshall only applies to directed acyclic graphs. So, instead of using Floyd-Warshall, I designed a solution around breadth-first search. When I discovered that Floyd-Warshall only fails on graphs that have negative cycles, I implemented a solution that employs Floyd-Warshall.
Even though the challenge is advertised as hard, I found it to be very easy. You can just recursively construct cliques and determine whether they are maximum. Although this is an exhaustive search, you can’t really solve the problem any faster than that, because the maximum clique problem is NP-complete. Fortunately, the size of the given problem instance is small enough for brute force to quickly yield a solution.
I solved today’s /r/DailyProgrammer challenge. The challenge is easy, but both of the optional bonus problems are significantly more difficult, and I solved both of them.
The first bonus problem is to decide whether any given grid of integers forms a magic square. The second bonus problem is, given a grid whose bottom row is missing, to decide whether the given grid could possibly form a magic square.
I was reading John D. Cook’s blog, and I read that all 3 x 3 magic squares have an interesting property. Being both curious and skeptical, I wrote a little Python program to see for myself whether it’s true that they really do. The code is up at my GitHub.
I have always liked dynamic programming. I went on a kick recently where I wrote solutions to several classic dynamic programming problems.
GeeksforGeeks published my implementation of the weighed interval scheduling problem. I wrote a Python program and a Java program to solve the weighted interval scheduling problem. When I was researching the problem, I noticed that the solution on GeeksforGeeks ran in O(n ^ 2) time, but I knew that the problem can be solved in O(n log n) time. So, I submitted a modification, and GeeksforGeeks accepted it.
Introduction to Algorithms, by Cormen, Leiserson, Rivest, and Stein, shows the optimal way to cut up a rod whose pieces you want to sell, and I implemented their solution in Python.
Bellman, Richard. Dynamic Programming. Princeton, NJ: Princeton University Press, 1957.