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);

We already know how we can find the shortest paths in a graph starting from a given vertex. Practically we modified breadth-first search in order to calculate the distances from s to all other nodes reachable from s. We know that this works because BFS walks through the graph level by level.

Some sources give a very simple explanation of how BFS finds the shortest paths in a graph. We must just think of the graph as a set of balls connected through strings.

As we can see by lifting the ball called “S” all other balls fall down. The closest balls are directly connected to “s” and this is the first level, while the outermost balls are those with longest paths.

Clearly edges like those between A and B doesn’t matter for our BFS algorithm because they don’t make the path from S to C through B shorter. This is also known as the triangle inequality, where the sum of the lengths of two of the sides of the triangle is always greater than the length of the third side.

We must only answer the question is BFS the best algorithm that finds the shortest path between any two nodes of the graph? This is a reasonable question because as we know by using BFS we don’t find only the shortest path between given vertices i and j, but we also get the shortest paths between i and all other vertices of G. This is an information that we actually don’t need, but can we find the shortest path between i and j without that info?

The answer is simply “no”! Practically depth-first search can’t help us. Even worse – we can find paths that are far not the shortest ones.

In the image above using DFS the distance between 1 and 7 is 7 while practically there is an edge between them.

So BFS is the optimal algorithm for finding shortest paths in a graph. But there’s a catch! This algorithm works fine when we assume that all the edges are the same length. In the examples so far each edge has the value of 1. So N edges between s and i made the distance between them of a length N.

Overview

As we know in practice different edges can have different values. Exactly that was the case in weighted graphs. Going back to the road map example the distances between different cities are commonly evaluated in miles or kilometers. Of course we can associate any other meaningful value to this edges. This can be either time in hours to travel between cities, money for fuel or anything else.

Now BFS can’t help us any more. Why? Because using non-equal values for the edges the triangle inequality is no longer true. Now the edge (the direct path) between A and B can be greater than the sum of the two edges (A, C) + (C, B)!

In other words, assuming the same abstraction with balls and wires the hanging wires can’t be discarded so easily.

So now how can we solve this problem? A very dummy approach is to break apart each edge with dummy vertices in order to make BFS work again.

However this approach has several weak points. The major one is that we’ll have to keep much more information, which means more memory usage, for even small graphs. This is done in case we break each edge on too many parts.

The solution of this problem was given by Edsger Dijkstra in 1956 and published in 1959. The only thing we should do now is to be sure that even discarding the triangle inequality we have the shortest paths. The first thing to do is to keep information for the distance from s to the parent (previous) node of i in the graph in order to calculate which distance is shorter.

In BFS we used a queue in order to walk through all the ancestors of a node. This was made consecutively. Thus for the graph G on the next image the order of enqueuing the ancestors of S was A, B, C.

The Dijkstra’s algorithm make use of a priority queue, also know as a heap. This fact combined by the fact we keep info for the shortest path so far help us find shortest paths in a weighted graphs.

Why this works? To answer this question let’s see the next very basic example, assuming the graph G from the next image. As we can see the triangle inequality isn’t true.

OK, we see that the path [S, B, A] is shorter than [S, A] although the edge (S, A) exists. How the Dijkstra algorithm overcomes this problem.

First we have no information about the distances (S, A) and (S, B), the only thing we know is that S is the starting point, its distance is 0 and its path so far is the empty set. So first we enqueue in a priority the distances from S to A and B.

Now we dequeue the minimum (first in the heap) element from the queue – the closest node to S, which is B. Then all the nodes adjacent to S in the queue are tested for adjacency to B, thus if we have already the distance between S and A now we can test if its longer than (S, B) + (B, A) – the triangle inequality!

So far we know that we must change a bit BFS to get the Dijkstra algorithm. The only thing to do is to keep info for each node for the path through its parent and to use a priority queue.

Code

Implementing this algorithms isn’t much more difficult than BFS, so here’s the code in PHP. However this example make use of the standard php library SPL and the PriorityQueue data structure, but any developer can code his own heap.

Here’s the graph from the code:

class vertex
{
    public $key         = null;
    public $visited     = 0;
    public $distance    = 1000000;  // infinite
    public $parent      = null;
    public $path        = null;
    
    public function __construct($key) 
    {
        $this->key  = $key;
    }
}

class PriorityQueue extends SplPriorityQueue
{
    public function compare($a, $b)
    {
        if ($a === $b) return 0;
        return $a > $b ? -1 : 1;
    }
}

$v0 = new vertex(0);
$v1 = new vertex(1);
$v2 = new vertex(2);
$v3 = new vertex(3);
$v4 = new vertex(4);
$v5 = new vertex(5);

$list0 = new SplDoublyLinkedList();
$list0->push(array('vertex' => $v1, 'distance' => 3));
$list0->push(array('vertex' => $v3, 'distance' => 1));
$list0->rewind();

$list1 = new SplDoublyLinkedList();
$list1->push(array('vertex' => $v0, 'distance' => 3));
$list1->push(array('vertex' => $v2, 'distance' => 7));
$list1->rewind();

$list2 = new SplDoublyLinkedList();
$list2->push(array('vertex' => $v1, 'distance' => 7));
$list2->push(array('vertex' => $v3, 'distance' => 8));
$list2->push(array('vertex' => $v4, 'distance' => 12));
$list2->rewind();

$list3 = new SplDoublyLinkedList();
$list3->push(array('vertex' => $v0, 'distance' => 1));
$list3->push(array('vertex' => $v2, 'distance' => 8));
$list3->rewind();

$list4 = new SplDoublyLinkedList();
$list4->push(array('vertex' => $v2, 'distance' => 12));
$list4->push(array('vertex' => $v5, 'distance' => 3));
$list4->rewind();

$list5 = new SplDoublyLinkedList();
$list5->push(array('vertex' => $v4, 'distance' => 3));
$list5->rewind();

$adjacencyList = array(
    $list0,
    $list1,
    $list2,
    $list3,
    $list4,
    $list5,
);

function calcShortestPaths(vertex $start, &$adjLists)
{
    // define an empty queue
    $q = new PriorityQueue();
    
    // push the starting vertex into the queue
    $q->insert($start, 0);
    $q->rewind();
    
    // mark the distance to it 0
    $start->distance = 0;
    
    // the path to the starting vertex
    $start->path = array($start->key);
    
    while ($q->valid()) {
        $t = $q->extract();
        $t->visited = 1;
        
        $l = $adjLists[$t->key];
        while ($l->valid()) {
            $item = $l->current();
            
            if (!$item['vertex']->visited) {
                if ($item['vertex']->distance > $t->distance + $item['distance']) {
                    $item['vertex']->distance = $t->distance + $item['distance'];
                    $item['vertex']->parent = $t;
                }
                
                $item['vertex']->path = array_merge($t->path, array($item['vertex']->key));
                
                $q->insert($item["vertex"], $item["vertex"]->distance);
            }
            $l->next();
        }
        $q->recoverFromCorruption();
        $q->rewind();
    }
}

calcShortestPaths($v0, $adjacencyList);

// The path from node 0 to node 5
// [0, 1, 2, 4, 5]
echo '[' . implode(', ', $v5->path) . ']';

Complexity

The complexity of that code is based on the complexity of BFS with the main difference that we keep a priority queue. For BFS we knew that the complexity was O(|V| + |E|), while Dijkstra’s algorithm has running time of O((|V| + |E|).log(|V|)). That is quite natural since the heapsort’s complexity is O(n.log(n))!

Application

Since the basic BFS can’t help us for weighted graphs and there are plenty of problems designed with weighted graphs obviously Dijkstra’s algorithm can be very handy. The only thing we should be aware of is the positive values of the edges.

Since with graphs we can represent real-life problems it’s almost clear why we would need an efficient algorithm that calculates the shortest path between two vertices. Getting back to our example of a road map we can use such an algorithm in order to find the shortest path between two cities. This example, of course, is very basic indeed, but it can give us a clear example of where shortest path can be applied.

In the other hand, we can model an enormous field of real-life problems using graphs – not only road maps. As we already know, whenever we have relations between different abstract objects we can refer an efficient graph algorithm.

OK, so we need a shortest path algorithm, but before we proceed with the exact algorithm first we’ll need to answer some questions and give some definitions.

Overview

First we need a definition of the terms distance and path between two nodes. A path is considered to be the sequence of vertices (or edges if you wish) between two vertices i and j. Of course we assume that there might be no path between any to vertices in the graph! Also we assume that this definition relates both for directed and undirected graphs. After we have the definition of a path we can proceed by defining a “distance”, which is said to be the number of edges in the path between i and j.

Using this terms, if there’s an edge between i and j, the path between them is [i, j], while the distance is 1. Of course, for an undirected graph (i, j) equals to (j, i) and the path [i, j] equals the path [j, i], but that isn’t true for directed graphs where the path [i, j] differs in general from [j, i].

Here we talk about the path between two adjacent vertices, but we can go with the more general case of a path between two vertices that aren’t adjacent.

Now, getting back to the road map example, there might be many paths between city A and city B.

What we actually need to find is the shortest one. This can be very important, because we often want to get from A to B as quickly as possible using the shortest path.

So first, what is a shortest path between i and j. Well, besides the strict definition, I’ll give a simplified one that might be clearer. The shortest path between i and j is such a path, which has the lowest distance compared to all other paths between i and j.

In our algorithm we will use breadth-first search. Why? That is because by using BFS by starting at a given point we expand our search consecutively starting with the closest vertices.

Is breadth-first search enough and will it give us the correct answer – the shortest path between i and j. Actually breadth-first search will gives us even more – the shortest paths to each reachable vertex from a given starting point – the staring vertex.

Why this is correct? Well, because of the nature of the breadth-first search algorithm. As we already know BFS uses a queue in order to store the front of the expansion. Usually as an abstraction BFS colors the vertices in white, gray and black, where the white vertices are those that aren’t visited yet, the gray are in the queue and the black vertices are already visited.

However how can be sure that BFS will give us the shortest paths to each vertex? To answer this question and to be sure that BFS will work for us we must take a closer look at the queue. Clearly by starting at a given point the algorithm is correct – the distance is 0.

Now the second step is to put into the queue all the vertices adjacent to s (where s is the starting point). Clearly this will give us the shortest paths to all adjacent vertices of s.

Continuing by induction we can assume that at level k we have all the shortest paths from s to all the vertices at the level k. It is clear the path between s and the vertices at level k is k, since we assume that each edge adds 1 to the path from s to i. Now by adding all the vertices adjacent (and not visited yet) to the paths of level k we get paths with length k+1 which is again the shortest paths from s to level k+1.

Actually we can talk about a tree built out of the graph by staring at s (which is the root of the tree).

Code

OK, now we know that BFS will find us the shortest paths from s to all the reachable vertices from s. Here’s a simple PHP implementation, that makes use of the Standard PHP Library data structures. Of course, everyone can code and use his own implementation of lists in order to keep the information of the adjacency lists.

The important thing to note is that we keep an additional information in each vertex – the distance between it and s, which is initially infinite. First we go with the modification of BFS in order to find all the distances between s and the other vertices.

Here’s our graph:

class vertex
{
    public $key = null;
    public $color = 'white';
    public $distance = -1;  // infinite
    
    public function __construct($key) 
    {
        $this->key = $key;
    }
}

$v0 = new vertex(0);
$v1 = new vertex(1);
$v2 = new vertex(2);
$v3 = new vertex(3);
$v4 = new vertex(4);
$v5 = new vertex(5);

$list0 = new SplDoublyLinkedList();
$list0->push($v1);
$list0->push($v3);
$list0->rewind();

$list1 = new SplDoublyLinkedList();
$list1->push($v0);
$list1->push($v2);
$list1->rewind();

$list2 = new SplDoublyLinkedList();
$list2->push($v1);
$list2->push($v3);
$list2->push($v4);
$list2->rewind();

$list3 = new SplDoublyLinkedList();
$list3->push($v1);
$list3->push($v2);
$list3->rewind();

$list4 = new SplDoublyLinkedList();
$list4->push($v2);
$list4->push($v5);
$list4->rewind();

$list5 = new SplDoublyLinkedList();
$list5->push($v4);
$list5->rewind();

$adjacencyList = array(
    $list0,
    $list1,
    $list2,
    $list3,
    $list4,
    $list5,
);

function calcDistances(vertex $start, &$adjLists)
{
    // define an empty queue
    $q = array();
    
    // push the starting vertex into the queue
    array_push($q, $start);
    
    // color it gray
    $start->color = 'gray';
    
    // mark the distance to it 0
    $start->distance = 0;
    
    while ($q) {
        // 1. pop from the queue
        $t = array_pop($q);
        
        // 2. foreach poped item find it's adjacent white vertices
        $l = $adjLists[$t->key];
        while ($l->valid()) {
            // 3. mark them gray, increment their length with one from their parent
            if ($l->current()->color == 'white') {
                $l->current()->color = 'gray';
                $l->current()->distance = $t->distance + 1;
                // 4. push them to the queue
                array_push($q, $l->current());
            }
            
            $l->next();
        }
    }
}

calcDistances($v0, $adjacencyList);

print_r($adjacencyList);

Now we can modify the algorithm even more and we add the path property of each vertex. Now each vertex will keep the path from s.

class vertex
{
    public $key         = null;
    public $color       = 'white';
    public $distance    = -1;  // infinite
    public $path        = null;
    
    public function __construct($key) 
    {
        $this->key  = $key;
    }
}

$v0 = new vertex(0);
$v1 = new vertex(1);
$v2 = new vertex(2);
$v3 = new vertex(3);
$v4 = new vertex(4);
$v5 = new vertex(5);

$list0 = new SplDoublyLinkedList();
$list0->push($v1);
$list0->push($v3);
$list0->rewind();

$list1 = new SplDoublyLinkedList();
$list1->push($v0);
$list1->push($v2);
$list1->rewind();

$list2 = new SplDoublyLinkedList();
$list2->push($v1);
$list2->push($v3);
$list2->push($v4);
$list2->rewind();

$list3 = new SplDoublyLinkedList();
$list3->push($v1);
$list3->push($v2);
$list3->rewind();

$list4 = new SplDoublyLinkedList();
$list4->push($v2);
$list4->push($v5);
$list4->rewind();

$list5 = new SplDoublyLinkedList();
$list5->push($v4);
$list5->rewind();

$adjacencyList = array(
    $list0,
    $list1,
    $list2,
    $list3,
    $list4,
    $list5,
);

function calcShortestPaths(vertex $start, &$adjLists)
{
    // define an empty queue
    $q = array();
    
    // push the starting vertex into the queue
    array_push($q, $start);
    
    // color it gray
    $start->color = 'gray';
    
    // mark the distance to it 0
    $start->distance = 0;
    
    // the path to the starting vertex
    $start->path = new SplDoublyLinkedList();
    $start->path->push($start->key);
    
    while ($q) {
        // 1. pop from the queue
        $t = array_pop($q);
        
        // 2. foreach poped item find it's adjacent white vertices
        $l = $adjLists[$t->key];
        while ($l->valid()) {
            // 3. mark them gray, increment their length with one from their parent
            if ($l->current()->color == 'white') {
                $l->current()->color = 'gray';
                $l->current()->distance = $t->distance + 1;
                $l->current()->path = clone $t->path;
                $l->current()->path->push($l->current()->key);
                
                // 4. push them to the queue
                array_push($q, $l->current());
            }
            
            $l->next();
        }
    }
}

calcShortestPaths($v0, $adjacencyList);

print_r($adjacencyList);

Complexity

Clearly the complexity of enqueue and dequeue is O(V), while searching for adjacent vertices is O(E), thus the complexity of this algorithm is O(V + E)!

Application

Finding the shortest path between two nodes is obviousely a very handy algorithm. Applied almost everywhere graphs exists this algorithm is widely used. However there’s one very reasonable question. We’re searching for the shortest path between two vertices and we end with the shortest paths between a starting node an all other vertices? Why we need this “useless” information? Acutally the question should be: is there a faster and more efficient algorithm compared to this one. Well, we’ll see that!