Tag Archives: programming

Floyd-Warshall Shortest Paths

The Floyd Warshall algorithm is an all pairs shortest paths algorithm. This can be contrasted with algorithms like Dijkstra’s which give the shortest paths from a single node to all other nodes in the graph.

Floyd Warshall’s algorithm works by considering first the edge set of the graph. This is the set of all paths of the graph through one edge. Node pairs that are connected to one another through an edge will have their shortest path set to the length of that edge, while all other node pairs will have their shortest path set to infinity. The program then runs through every triplet of nodes (i, j, k) and checks if the path from i to k and the path from k to j is shorter than the current path from i to j. If so, then the distance and the path is updated.

So lets consider an example on the graph in the image above. The edge set of this graph is E = {(0, 1), (0, 2), (0, 3), (1, 3), (3, 4)}. So our initial table is:

  0 1 2 3 4
0 inf (0, 1) (0, 2) (0, 3) inf
1 (0, 1) inf inf (1, 3) inf
2 (0, 2) inf inf inf inf
3 (0, 3) (1, 3) inf inf (3, 4)
4 inf inf inf (3, 4) inf

As we look to update the paths, we first look for routes that go through node 0:

Because node 0 connects to both node 1 and node 2, but node 1 does not connect to node 2, we have the following truth holding in the matrix above:
cost(0, 1) + cost(0, 2) < cost(1, 2), so we can update the shortest path from node 1 to node 2 to be (1, 0, 2).

Because node 0 connects to both node 2 and node 3, but node 2 does not connect to node 3, we have the following truth holding in the matrix above:
cost(0, 2) + cost(0, 3) < cost(2, 3), so we can update the shortest path from node 2 to node 3 to be (2, 0, 3).

Because node 3 connects to both node 0 and node 4, but node 0 does not connect to node 4, we have the following truth holding in the matrix above:
cost(0, 3) + cost(3, 4) < cost(0, 4), so we can update the shortest path from node 0 to node 4 to be (0, 3, 4).

Because node 3 connects to both node 1 and node 4, but node 1 does not connect to node 4, we have the following truth holding in the matrix above:
cost(1, 3) + cost(3, 4) < cost(1, 4), so we can update the shortest path from node 1 to node 4 to be (1, 3, 4).

Because node 3 connects to both node 2 and node 4, but node 2 does not connect to node 4, we have the following truth now holding:
cost(2, 3) + cost(3, 4) < cost(2, 4), so we can update the shortest path from node 2 to node 4 to be (2, 0, 3, 4).

The final table giving the list of shortest paths from every node to every other node is given below.

  0 1 2 3 4
0 inf (0, 1) (0, 2) (0, 3) (0, 3, 4)
1 (0, 1) inf (1, 0, 2) (1, 3) (1, 3, 4)
2 (0, 2) (1, 0, 2) inf (2, 0, 3) (2, 0, 3, 4)
3 (0, 3) (1, 3) (2, 0, 3) inf (3, 4)
4 (0, 3, 4) (1, 3, 4) (2, 0, 3, 4) (3, 4) inf

To see more examples and to help answer questions, check out the script in my examples section on the Floyd-Warshall algorithm

Degree Centrality of a Graph

Degree Centrality Example

I wanted to spend some time on centrality measures of a graph. These are measurements of how important each node (or edge) is to the overall graph. But how do we define, or determine, importance? There is no unique way to answer this question, so there are varying metrics for measuring centrality. Which one you choose depends on several factors including how many other nodes of the graph are included, as well as the run time of the metrics you’re considering.

I have just published a script focusing on the degree centrality metric. The degree centrality metric is called a “walk metric” because it determines how important a node is by how many other nodes that can be reached by walks of up to a certain length. Lets look at the definition of the degree of a node to see if we can understand why it is called a walk metric.

In an undirected graph G = (V, E), the degree of a node u [in] V is the |{v | (u, v) [in] E}|. This is the size of the set of nodes that are connected to node u via a single edge. Another way of describing a single edge is a walk of length one. So the degree metric measures the importance of a node by the number of unique walks of length one.

The normalized degree centrality of a node v in a graph G = (V,E) measures how many nodes are connected to the node v, compared to the maximum possible number of edges that can be connected to this node. Because we are dealing with simple undirected graphs (at most a single edge between any two distinct vertices), this maximum possible number will always be |V – 1|. So the normalized degree can be calculated by dividing the degree of the node (the number of nodes it is connected to) by |V – 1|.

So for the example above, the node 0 has degree 6 because it is connected to nodes 2, 5, 9, 10, 11, and 12. There are 15 total nodes in this graph, so to calculate the normalized degree centrality of the node 0, it will be 6 / 14, which rounds to 0.428571.

To see more examples and to help answer questions, check out the script in my examples section on degree centrality

Tarjan’s Strongly Connected Components Algorithm

I just added a program that finds the strongly connected components of a graph using Tarjan’s Algorithm.

A strongly connected component of a graph is a subgraph S of G where every pair of nodes, u and v in S there is a path from u to v and a path from v to u.

To find these strongly connected components we implement Tarjan’s algorithm. The idea behind Tarjan’s algorithm is to begin by running a depth first search from an arbitrary node in the graph, labeling nodes reachable from this start node in the order they are reached. The algorithm is also interested in the “oldest” node that could be reached by a given node. This is indicated by the keeping track of the lowest label that can be reached from that node. We will call the first property label(v) and the second lowlink(v).

When the algorithm starts label(v) is the same as lowlink(v) whenever a node is discovered. As the algorithm is executed, the DFS is being run on each discovered node, which in turn updates the lowlink(v) property telling of (older) nodes that can be reached. If an older node can be reached, then we update lowlink. If we reach a node that cannot connect to any older nodes after the DFS call, i.e if label(v) is the same a lowlink(v), then this means that this node does not have a path to any node with a lower label. So this node will be the first node of a new strongly connected component.

Feel free to check it out an let me know what you think in the comments below.

Longest Common Subsequence

Suppose you and I each had an ordered list of items and we were interested in comparing how similar those lists are. One calculation we can perform on these two strings is the Longest Common Subsequence. A sequence X is an ordered list of elements <x1, …, xn>. A subsequence Z is another sequence where (1) Each element of Z is also an element of X and (2) The elements of Z occur in the same order (in Z) as they do in X.

Note that we do not say that the elements of Z need to be a continuous block of elements. If this were true we would be defining a substring. So as an example, suppose we have as an initial string,
X = C, O, M, P, U, T, E, R. 
Then the following are all subsequences: 
Z1 = C, M, U, T, R
Z2 = C, O, M, P
Z3 = U, T, E, R
Z4 = O, P, T, E

I will note that Z2 and Z3 are also substrings since they contain continuous sets of characters. 

The length of a substring is simply the number of characters it contains. So X has length 8, Z1 has length 5, Z2, Z3 and Z4 have length 4. 

Suppose now that we had a second string, Y = P, R, O, G, R, A, M, M, E, R and are interested in the longest common subsequence between the two. We can do that by observing that there is a bit of recursion going on with this question. What I mean by that is that asking the question of “What is the longest common subsequence between X and Y” is the same as asking “What is the longest common subsequence between X and Y once we have seen 8 characters of X and 10 characters of Y”

There are three possible ways to answer this question. 

If X<sub>8</sub> equals Y<sub>10</sub>, then we ask the same question about X<sub>7 and Y<sub>9</sub> and add 1 to the answer. 
If X<sub>8</sub> is not equal to Y<sub>10</sub>, then the answer to this will be the same as the maximum of the pair X<sub>7</sub>, Y<sub>10</sub> and the pair X<sub>8</sub>, Y<sub>9</sub>. 
If we reach a situation where we reach the beginning of either string, we are forced to answer 0 to that question. 

Then the function has the following look: 

LCS(Xi, Yj) =
0, if i is 0 or j is 0
1 + LCS(Xi-1, Yj-1) if Xi equals Yj
max(LCS(Xi-1, Yj), LCS(Xi, Yj-1))

Below is a table showing how we would solve the problem mentioned.

The strategy used to devise this concept is called dynamic programming. It is useful we can solve larger problems by solving overlapping subproblems, as was the case here. In this situation we generally can store the data in a table form and avoid re-solving subproblems for which many larger problems will be dependent.

You can see this algorithm in action at LEARNINGlover.com: Longest Common Subsequece. Let me know what you think.

Independent Set Puzzles

In this post, I want to return to the idea of NP-Complete problems. There is a more technical, more formal definition that I can refer you to, but I like to refer to the images from Garey and Johnson’s “Computers and Intractability: A Guide to the Theory of NP Completeness”. The images helps to understand the difficulty of NP-Complete problems by presenting two images. The first image shows a single individual speaking to someone saying that he has been unable to solve the problem. The second image shows that same individual speaking to the same person behind the desk, but saying that not only was he unable to solve the problem but neither was a long line of people. The theory of NP Complete problems revolves around the concept that if an efficient algorithm exists for an NP-Complete problem, then an efficient algorithm exists for all problems in the class NP.

Today, I would like to present a puzzle I created to play with the Independent Set problem. This is the problem where we are given a graph G = (V, E) and are asked to find a maximum set of vertices S such that there is no edge in the graph G between any two vertices in S. The decision version (the problem asking whether there is an independent set of size k) of this problem is NP-Complete, so the known algorithms for problem either have a slow running time, or do not solve it exactly.

This problem is very related to another puzzle I posted last year called the clique problem. In fact, Karp originally proved that Clique was NP-Complete by showing that if Clique could be solved efficiently, then Independent Set could be solved efficiently. He did this by constructing a second graph [G bar], called the compliment of G (containing the same vertices in G, along with the edges that are not present in G. Any edge present in G will not be present in ). Then he showed that the nodes representing a maximum clique in G would represent a maximum independent set in . He had already shown that Independent Set was NP-Complete, which meant that both Independent Set and Clique were among the most difficult problems within the class known as NP.

The puzzle begins with an undirected graph and asks users to find a maximum independent set. Users should click on the numbers in the table below the graph indicating the nodes they wish to select in their independent set (purple indicates that the node is selected, gray indicates that it is not). Once a user have a potential solution, they can press the “Check” button to see if their solution is optimal. If a user is having trouble and simply wishes to see the maximum independent set, they can press the “Solve” button. And to generate a new problem, users can press the “New Problem” button.As a result of this relationship between the Clique problem and the Independent Set problem, the Bron-Kerbosch algorithm that was used to find maximum cliques previously can also be used here.

The Depth-First-Search Algorithm

I remember when I was younger I used to play the game of hide-and-seek a lot. This is a game where a group of people (at least two) separate into a group of hiders and a group of seekers. The most common version of this that I’ve seen is having one person as the seeker and everyone as hiders. Initially, the seeker(s) is given a number to count towards and close their eyes while counting. The hiders then search for places to hide from the seeker. Once the seeker is finished counting, their job is to find where everyone is hiding or admitting that they cannot find all the seekers. Any seekers not found are said to have won, and seekers that are found are said to have lost.

I played this game a number of times in my childhood, but I remember playing it with a friend named Dennis in particular. Dennis had a certain way he played as seeker. While many of us would simply go to places we deemed as “likely” hiding spots in a somewhat random order, Dennis would always begin by looking in one area of the room, making sure that he had searched through every area connected to that area before going to a new area. He continued this process until he either found everybody or concluded that he had searched every spot he could think of and gave up.

It wasn’t until years later that I was able to note the similarity between Dennis’s way of playing hide-and-seek and the Depth-First-Search algorithm. The Depth-First-Search Algorithm is a way of exploring all the nodes in a graph. Similar to hide-and-seek, one could choose to do this in a number of different ways. Depth-First-Search does this by beginning at some node, looking first at one of the neighbors of that node, then looking at one of the neighbors of this new node. If the new node does not have any new neighbors, then the algorithm goes to the previous node, looks at the next neighbor of this node and continues from there. Initially all nodes are “unmarked” and the algorithm proceeds by marking nodes as being in one of three states: visited nodes are marked as “visited”; nodes that we’ve marked to visit, but have not visited yet are marked “to-visit”; and unmarked nodes that have not been marked or visited are “unvisited”.

Consider a bedroom with the following possible hiding locations: (1) Under Bed, (2) Behind Cabinet, (3) In Closet, (4) Under Clothes, (5) Behind Curtains, (6) Behind Bookshelf, and (7) Under Desk. We can visualize how the bedroom is arranged as a graph and then use a Breadth First Search algorithm to show how Brent would search the room. Consider the following bedroom arrangement, where we have replaced the names of each item by the number corresponding to that item. Node (0) corresponds to the door, which is where Dennis stands and counts while others hide.

Bedroom Items as a Graph

Now consider how a Breadth First Search would be run on this graph.

Bedroom Items as a Graph Colored by DFS

The colors correspond to the order in which nodes are visited in Depth-First-Search.

The way we read this is that initially Dennis would start at node 0, which is colored in Blue.
While Dennis is at node 0, she notices that nodes 1, 5, and 6 (under bed, behind curtains, and behind bookshelf) are the nearby and have not been checked yet so she places them on the “to visit” list.
Next, Dennis will begin to visit each node on the “to visit” list, and when a node is visited, she labels it as visited. At each location, she also takes note of the other locations she can reach from this location. Below is the order of nodes Dennis visits and how he discovers new locations to visit.

Order Visited Node Queue Adding Distance From Node 0
1 0 6,5,1 0
2 6 5,1 7,3,2 1
3 7 3,2,5,1 2
4 3 2,5,1 2
5 2 5,1 4 2
6 4 5,1 3
7 5 1 1
8 1 1

Here is a link to my Examples page that implements the Depth-First-Search Algorithm on Arbitrary Graphs.

The Breadth-First-Search Algorithm

I remember when I was younger I used to play the game of hide-and-seek a lot. This is a game where a group of people (at least two) separate into a group of hiders and a group of seekers. The most common version of this that I’ve seen is having one person as the seeker and everyone as hiders. Initially, the seeker(s) is given a number to count towards and close their eyes while counting. The hiders then search for places to hide from the seeker. Once the seeker is finished counting, their job is to find where everyone is hiding or admitting that they cannot find all the seekers. Any seekers not found are said to have won, and seekers that are found are said to have lost.

I played this game a number of times in my childhood, but I remember playing it with a friend named Brenda in particular. Brenda had a certain way she played as seeker. While many of us would simply go to places we deemed as “likely” hiding spots in a somewhat random order, Brenda would always take a survey of the room, and no matter where she began searching, she would always make note of the locations close to her starting point and make sure she was able to give them all a look before she looked at locations that were close to the points she deemed close to the starting point. She continued this process until she either found everybody or concluded that she had searched every spot she could think of and gave up.

It wasn’t until years later that I was able to note the similarity between Brenda’s way of playing hide-and-seek and the Breadth-First-Search algorithm. The Breadth-First-Search algorithm is a way of exploring all the nodes in a graph. Similarly to hide-and-seek, one could choose to do this in a number of different ways. Breadth-First-Search does this by beginning at some node, looking first at each of the neighbors of the starting node, then looking at each of the neighbors of the neighbors of the starting node, continuing this process until there are no remaining nodes to visit. Initially all nodes are “unmarked” and the algorithm proceeds by marking nodes as being in one of three stages: visited nodes are marked as “visited”; nodes that we’ve marked to visit, but have not visited yet are marked “to-visit”; and unmakred nodes that have not been marked are “unvisited”.

Consider a bedroom with the following possible hiding locations: (1) Under Bed, (2) Behind Cabinet, (3) In Closet, (4) Under Clothes, (5) Behind Curtains, (6) Behind Bookshelf, and (7) Under Desk. We can visualize how the bedroom is arranged as a graph and then use a Breadth First Search algorithm to show how Brenda would search the room. Consider the following bedroom arrangement, where we have replaced the names of each item by the number corresponding to that item. Node (0) corresponds to the door, which is where Brenda stands and counts while others hide.

Bedroom Items as a Graph

Now consider how a Breadth First Search would be run on this graph.

Bedroom Items as a Graph

The colors correspond to the order in which nodes are visited in Breadth-First-Search.

The way we read this is that initially Brenda would start at node 0, which is colored in Blue.
While Brenda is at node 0, she notices that nodes 1, 5, and 6 (under bed, behind curtains, and behind bookshelf) are the nearby and have not been checked yet so she places them on the “to visit” list.
Next, Brenda will begin to visit each node on the “to visit” list, and when a node is visited, she labels it as visited. At each location, she also takes note of the other locations she can reach from this location. Below is the order of nodes Brenda visits and how she discovers new locations to visit.

Order Visited Node Queue Adding Distance From Node 0
1 0 1, 5, 6 0
2 1 5, 6 2, 4 1
3 5 6, 2, 4 1
4 6 2, 4 3, 7 1
5 2 4, 3, 7 2
6 4 3, 7 2
7 3 7 2
8 7 2

Here is a link to my Examples page that implements the Breadth-First-Search Algorithm on Arbitrary Graphs.

Clique Problem Puzzles

I still remember how I felt when I was first introduced to NP-Complete problems. Unlike the material I had learned up to that point, there seemed to be such mystery and intrigue and opportunity surrounding these problems. To use the example from Garey and Johnson’s book “Computers and Intractability: A Guide to the Theory of NP Completeness”, these were problems that not just one researcher found difficult, but that a number of researchers had been unable to find efficient algorithms to solve them. So what they did was show that the problems all had a special relationship with one another, and thus through this relationship if someone were to discover an algorithm to efficiently solve any one of these problems they would be able to efficiently solve all the problems in this class. This immediately got my mind working into a world where I, as a college student, would discover such an algorithm and be mentioned with the heavyweights of computer science like Lovelace, Babbage, Church, Turing, Cook, Karp and Dean.

Unfortunately I was a student so I did not have as much time to devote to this task as I would have liked. In my spare time though I would try to look at problems and see what kind of structure I found. One of my favorite problems was, The Clique Problem. This is a problem where we are given an undirected graph and seek to find a maximum subset of nodes in this graph that all have edges between them, i.e. a clique (Actually the NP-Complete version of this problem takes as input an undirected graph G and an integer k and asks if there is a clique in G of size k).

Although I now am more of the mindset that there do not exist efficient algorithms to solve NP-Complete problems, I thought it would be a nice project to see if I could re-create this feeling – both in myself and others. So I decided to write a program that generates a random undirected graph and asks users to try to find a maximum clique. To test users answers, I coded up an algorithm that works pretty well on smaller graphs, the Bron-Kerbosch Algorithm. This algorithm uses backtracking to find all maximal cliques, which then allows us to sort them by size and determine the largest.

Users should click on the numbers in the table below the canvas indicating the nodes they wish to select in their clique (purple indicates that the node is selected, gray indicates that it is not). Once they have a potential solution, they can press the “Check” button to see if their solution is optimal. If a user is having trouble and simply wishes to see the maximum clique, they can press the “Solve” button. And to generate a new problem, users can press the “New Problem” button.

So I hope users have fun with the clique problem puzzles, and who knows maybe someone will discover an algorithm that efficiently solves this problem and become world famous.

Triangle Sum Puzzle

This is probably a consequence of being a mathematician, but I have always enjoyed number puzzles. I think that there is a general simplicity and universality in numbers that are not present in things like word puzzles, where the ability to reach a solution can be limited to the vocabulary of the user.

The fact that these are puzzles and not simply homework exercises also helps because we often find people sharing difficulties and successes stories over the water cooler or at the lunch table. The fact that many of these math puzzles can teach some of the same concepts as homework problems (in a more fun and inclusive way) is generally lost on the user as their primary interest is generally on solving the puzzle in front of them, or sometimes solving the more general form of the puzzle.

Today’s post is about a puzzle that was originally shared with me over a lunch table by a friend who thought it was an interesting problem and asked what I thought about it. I didn’t give the puzzle much further thought (he had correctly solved the puzzle) until I saw it again in “Algorithmic Puzzles” by Anany Levitin and Maria Levitin. It was then that I thought about the more general form of the puzzle, derived a solution for the problem, and decided to code it up as a script for my site.

Below is a link to the puzzle:

We have a set of random numbers arranged in a triangle and the question is to find the path of maximum sum from the root node (the top node) to the base (one of the nodes on the bottom row) with the rules that
(1) Exactly one number must be selected from each row
(2) A number can only be selected from a row if (a) it is the root node or (b) one of the two nodes above it has been selected.

For the sample
So for the sample problem in the picture, the maximal path would go through nodes 57, 99, 34, 95, and 27.

For more of these puzzles check out the script I write here and be sure to let me know what you think.

Hierarchical Clustering

Hierarchical Clustering algorithms give a nice introduction for computer science students to unsupervised machine learning. I say this because the bottom-up approach to Hierarchical clustering (which I have implemented here) is very similar to Kruskal’s algorithm for finding the minimum spanning tree of a graph.

In Kruskal’s algorithm, we begin by creating a forest, or a set of trees where each node is its own tree. The algorithm then selects the two trees that are closest together (closest being defined as the minimum cost edge between two distinct trees) and merges those trees together. This process of merging the closest two trees is then repeated until there is only one tree remaining, which is a minimum spanning tree of the graph.

Similarly, bottom-up hierarchical clustering of a group of points begins by saying that each point is its own cluster. Then the clusters are compared to one another to check if two clusters will be merged into one. Generally, there will be some stopping criteria, , saying that we do not want to merge two clusters together if their distance is greater than . So if the minimum distance between two clusters is less than we will proceed as in Kruskal’s algorithm by merging these two clusters together into one cluster. We repeat this process of merging the closest two clusters together until we find that the minimum distance between two clusters is greater than or equal to , in which case we can stop and the result is a partition of our data set into distinct clusters.

Hierarchical clustering is comparable to K-Means Clustering. Here are some differences between the two approaches:

  1. K-Means Clustering requires an initial number of desired clusters, while Hierarchical clustering does not.
  2. A run of K-Means Clustering will always give K clusters, whereas Hierarchical Clustering can give more or less, depending on our tolerance .
  3. K-Means can undo previous mistakes (assignments of an element to the wrong cluster), while Hierarchical Clustering cannot.

So, here is a link to my page on Hierarchical Clustering. Hope you enjoy.