A very common problem in computer programming is finding the longest increasing (decreasing) subsequence in a sequence of numbers (usually integers). Actually this is a typical dynamic programming problem.

Dynamic programming can be described as a huge area of computer science problems that can be categorized by the way they can be solved. Unlike divide and conquer, where we were able to merge the fairly equal sub-solutions in order to receive one single solution of the problem, in dynamic programming we usually try to find an optimal sub-solution and then grow it.

Once we have an optimal sub-solution on each step we try to upgrade it in order to cover the whole problem. Thus a typical member of the dynamic programming class is finding the longest subsequence.

However this problem is interesting because it can be related to graph theory. Let’s find out how.

Overview

We already know various ways to calculate the shortest paths in a graph. Indeed finding the single-source shortest path is a typical graph problem. To model such kind of solutions we definitely need a graph represented in our solution.

However the single-source shortest path isn’t a straight-forward problem. It depends on many factors. Thus for positive edges the algorithm of Edsger Dijkstra can be a perfect solution, but when the graph contains negative edges his algorithm is no longer useful.

In the presence of negative edges the Dijkstra’s algorithm doesn’t work and we’d better use the Bellman-Ford algorithm. It is interesting to note, that it was exactly Richard Bellman who first introduced the term “dynamic programming” in the 1940s.

In fact the Bellman-Ford algorithm was able to detect negative cycles. That’s too important, because in presence of negative cycles the shortest path problem is no longer well defined.

In the other hand when we’re talking about shortest paths in a DAG (Directed Acyclic Graph) we can find a faster (linear) solution. That’s because we’re sure that there are no cycles (not even negative cycles)!

Finding the shortest paths in a DAG is closely related to the topological sorting of the DAG. This gives us a linear representation of the vertices of the DAG and we can clearly calculate the distances from the starting node to all other nodes. Note that in a DAG we have one or more nodes that can be considered as starting nodes – which means they don’t have predecessors (incoming edges).

Following the words above is pretty hard to find what all these graph algorithms have to do with finding the longest increasing (decreasing) subsequence. Actually this problem is very closely related to the toplogical sort of the DAG and the problem of finding shortest paths in a DAG.

That’s because we can represent our sequence as a DAG. The only thing we must care about is to “connect” with directed edges those elements that form an increasing (decreasing) pair.

Thus the sequence from our example [1, 8, 2, 7, 3, 4, 1, 6] is going to look like this.

Another important thing to note is that we don’t search for shortest, but for longest path, since our task is to find the longest subsequence.

Pseudo Code

Our pseudo code for finding the shortest paths in a DAG was something like that.

1. Get the toplogically sorted list L of the DAG;
2. The starting node is s;
3. The distance to s equals to 0;
4. All other distances are initialized to ∞
5. For each node (v) in L\{s} do:
5.1. If dist(v) > dist(u) + w(u, v) then
5.1.1.  dist(v) := dist(u) + w(u, v)

In the above pseudo code “u” is every predecessor of “v”!

Now we must “reverse” the solution above in order to find the longest increasing subsequence. Note that we don’t care any more about the weight of the edges, thus we can simply substitute them with 1.

1. Get the sequence (L) as a toplogically sorted DAG;
2. For each (i) in L do:
2.1. S(i) := 1 + max(S(j), where (i, j) is an edge from the DAG);

Application

Finding the longest increasing subsequence can be very useful not only at the Google/Yahoo/Facebook interview, but also in various fields of statistics.

We saw how to find the shortest path in a graph with positive edges using the Dijkstra’s algorithm. We also know how to find the shortest paths from a given source node to all other nodes even when there are negative edges using the Bellman-Ford algorithm. Now we’ll see that there’s a faster algorithm running in linear time that can find the shortest paths from a given source node to all other reachable vertices in a directed acyclic graph, also known as a DAG.

Because the DAG is acyclic we don’t have to worry about negative cycles. As we already know it’s pointless to speak about shortest path in the presence of negative cycles because we can “loop” over these cycles and practically our path will become shorter and shorter.

Thus we have two problems to overcome with Dijkstra and the Bellman-Ford algorithms. First of all we needed only positive weights and on the second place we didn’t want cycles. Well, we can handle both cases in this algorithm.

Overview

The first thing we know about DAGs is that they can easily be topologically sorted. Topological sort can be used in many practical cases, but perhaps the mostly used one is when trying to schedule dependent tasks.

After a topological sort we end with a list of vertices of the DAG and we’re sure that if there’s an edge (u, v), u will precede v in the topologically sorted list.

This information is precious and the only thing we need to do is to pass through this sorted list and to calculate distances for a shortest paths just like the algorithm of Dijkstra.

OK, so let’s summarize this algorithm:
– First we must topologically sort the DAG;
– As a second step we set the distance to the source to 0 and infinity to all other vertices;
– Then for each vertex from the list we pass through all its neighbors and we check for shortest path;

It’s pretty much like the Dijkstra’s algorithm with the main difference that we used a priority queue then, while this time we use the list from the topological sort.

Code

This time the code is actually a pseudocode. Altough all the examples so far was in PHP, perhaps pseudocode is easier to understand and doesn’t bind you in a specific language implementation. Also if you don’t feel comforatable with the given programming language it can be more difficult for you to understand the code than by reading pseudocode.

1. Topologically sort G into L;
2. Set the distance to the source to 0;
3. Set the distances to all other vertices to infinity;
4. For each vertex u in L
5.    - Walk through all neighbors v of u;
6.    - If dist(v) > dist(u) + w(u, v) 
7.       - Set dist(v) <- dist(u) + w(u, v);

Application

It’s clear why and where we must use this algorithm. The only problem is that we must be sure that the graph doesn’t have cycles. However if we’re aware of how the graph is created we may have some additional information if there are cycles or not – then this linear time algorithm can be very applicable.

As we saw in the previous post, the algorithm of Dijkstra is very useful when it comes to find all the shortest paths in a weighted graph. However it has one major problem! Obviously it doesn’t work correctly when dealing with negative lengths of the edges.

We know that the algorithm works perfectly when it comes to positive edges, and that is absolutely normal because we try to optimize the inequality of the triangle.

Since Dijkstra’s algorithm make use of a priority queue normally we get first the shortest adjacent edge to the starting point. In our very basic example we’ll get first the edge with the length of 3 -> (S, A).

However when it comes to negative edges we can’t use any more priority queues, so we need a different, yet working solution.

Overview

The solution was published by Richard E. Bellman and Lester Ford, Jr. in 1958 in their publication “On a Routing Problem” and it is quite simple to explain and understand. Since we can prioritize the edges by its lengths the only thing we should do is to calculate all the paths. And to be sure that our algorithm will find all the paths correctly we repeat that N-1 times, where N is the number of vertices (|V| = N)!

In this very basic image we can see how Bellman-Ford solves the problem. First we get the distances from S to A and B, which are respectively 3 and 4, but there is a shorter path to A, which passes through B and it is (S, B) + (B, A) = 4 – 2 = 2.

Code

Here’s the code on PHP. Note that this time we use an adjacency matrix and an additional array of distances. It’s important (for directed graphs, and our graph this time is directed) to put the positive value of A[j][i] if A[i][j] is negative. Note the case for A[1][2]!

define('INFINITY', 10000000);

$matrix = array(
    0 => array( 0,  3,  4),
    1 => array( 0,  0,  2),
    2 => array( 0,  -2, 0),
);

$len = count($matrix);

$dist = array();

function BellmanFord(&$matrix, &$dist, $start)
{
    global $len;
    
    foreach (array_keys($matrix) as $vertex) {
        $dist[$vertex] = INFINITY;
        if ($vertex == $start) {
            $dist[$vertex] = 0;
        }
    }
    
    for ($k = 0; $k < $len - 1; $k++) {
        for ($i = 0; $i < $len; $i++) {
            for ($j = 0; $j < $len; $j++) {
                if ($dist[$i] > $dist[$j] + $matrix[$j][$i]) {
                    $dist[$i] = $dist[$j] + $matrix[$j][$i];
                }
            }
        }
    }
}

BellmanFord($matrix, $dist, 0);

// [0, 2, 4]
print_r($dist);

Complexity

The complexity is clearly O(n³) which follows directly from the code above.

Application

Actually this algorithm is very useful and it not only works with negative weights, but also can help us find negative cycles in the graph.

This is done with the simple check after the main loop.

    for ($i = 0; $i < $len; $i++) {
        for ($j = 0; $j < $len; $j++) {
            if ($dist[$i] > $dist[$j] + $matrix[$j][$i]) {
                echo 'The graph contains a negative cycle!';
            }
        }
    }

And here’s the full code.

$matrix = array(
    0 => array( 0,  3,  4),
    1 => array( 0,  0,  2),
    2 => array( 0,  -2, 0),
);

$len = count($matrix);

$dist = array();

function BellmanFord(&$matrix, &$dist, $start)
{
    global $len;
    
    foreach (array_keys($matrix) as $vertex) {
        $dist[$vertex] = INFINITY;
        if ($vertex == $start) {
            $dist[$vertex] = 0;
        }
    }
    
    for ($k = 0; $k < $len - 1; $k++) {
        for ($i = 0; $i < $len; $i++) {
            for ($j = 0; $j < $len; $j++) {
                if ($dist[$i] > $dist[$j] + $matrix[$j][$i]) {
                    $dist[$i] = $dist[$j] + $matrix[$j][$i];
                }
            }
        }
    }
    
    for ($i = 0; $i < $len; $i++) {
        for ($j = 0; $j < $len; $j++) {
            if ($dist[$i] > $dist[$j] + $matrix[$j][$i]) {
                echo 'The graph contains a negative cycle!';
            }
        }
    }
}

BellmanFord($matrix, $dist, 0);

// [0, 2, 4]
print_r($dist);