Perhaps right after the addition at school we’ve learned how to multiply two numbers. This algorithm isn’t as easy as addition, but besides that we’re so familiar with it and that we even don’t recognize it as an “algorithm”. We just know it by heart.

However, as I already said, multiplication is a bit more difficult than addition. This algorithm is interesting because for several reasons. First of all, let’s compare multiplication in binary and decimal.

So, let’s see how to multiply two numbers.

Overview

Multiplying and adding is practically the same thing, so where’s the difference?

It’s clear that a product of two numbers (each with N digits) can be represented as N sums, as we can see on the next picture.

Now, for larger integers each time we shift left the next intermediate sum.

That is absolutely logical since we can represent the numbers as a sum of decimals divisible by 10 without a remainder.

Binary Multiplication

Sometimes binary is easier to work with than decimal and multiplication is just the case. As shown on the picture below binary multiplication is much easier compared to decimal.

That’s because we multiply only by 1 and 0, so the intermediate sum can be either the first number or 0.

In the oder hand, shifting left in binary equals multiplication by 2.

Why? Well, simply because the base is 2. It’s practically the same with decimals where shifting left equals multiplying by 10.

The same applies of course for any base, i.e. hex F equals decimal 15 (since the first number is 0) and FF equals 255 (which is 16×16 – 1).

Can we do better?

Unlike addition, here the answer is – yes, and we already know how to multiply faster (either decimal or binary numbers) using the Karatsuba’s fast multiplication algorithm.

It’s Important Because …

I’ve to admit that both addition and multiplication algorithms are fairly to understand, but we must remember that they are giving us the ground level for more complex cryptographic algorithms.

We know how to add two integers using a perfectly simple and useful algorithm learned from school or even earlier. This is perhaps one of the very first techniques we learn in mathematics. However we need to answer few questions. First of all do computers use the same technique, since they use binary representation of numbers? Is there a faster algorithm used by computers? What about boundaries and large integers?

Overview

Let’s start by explaining how we humans add two numbers. An important fact is that by adding two single-digit numbers we get at most two digit number. This can be proven by simply realizing that 9+9 = 18. This fact lays down in the way we add integers. Here’s how.

We just line-up the integers on their right-most digit and we start adding them in a column. In case we got a sum greater than 9 (let’s say 14) we keep only the right-most digit (the 4) and the 1 is added to the next sum.

Thus we get to the simple fact that by adding two n-digit integers we can have either an n-digit integer or a n+1 digit integer. As an example we see that by adding 53 + 35 (two 2-digit integers) we get 88, which is again 2-digit integer, but 53 + 54 result in 107, which is 3 digit integer.

That fact is practically true, as I mentioned above, for each pair of n-digit integers.

What about binary numbers?

In fact binaries can be added by using the exact same algorithm. At the example below we add two integers represented as binary numbers.

As a matter of fact this algorithm is absolutely wonderful, because it works not only on decimals and binaries but in any base B.

Of course computers tend to perform better when adding integers that “fit” the machine word. However as we can see later this isn’t always the case and sometimes we need to add larger numbers that exceed the type boundaries.

What about big integers?

Since we know how to add “small” integers, it couldn’t be so hard to apply the same algorithm on big integers. The only problem is that the addition will be slower and sometimes (done by humans) can be error prone.

So practically the algorithm is the same, but we can’t just put a 1 billion integer into a standard computer type INT, right? That means that the tricky part here is the way we represent integers in our application. A common solution is to store the “big” integer into an array, thus each digit will be a separate array item. Then the operation of addition will be simple enough to be applied.

Complexity

When we talk about an algorithm that is so well known by every human being (or almost every) a common question is “is there anything faster” or “do computers use a different algorithm”. The answer may be surprising to someone, but unfortunately that is the fastest (optimal) algorithm for number addition.

Practically there’s nothing to optimize here. We just read the two n-digit numbers (O(n)), we apply “simple” addition to each digit and we carry over the 1 from the sums greater than 9 to the next “simple” addition. We don’t have loops or any complex operation in order to search for an optimization niche.

Application

It’s strange how often this algorithm is asked on coding interviews. Perhaps the catch is whether the interviewed person will start to look for a faster approach?! Thus is cool to know that this algorithm is optimal.

Sometimes we may ask ourselves why we humans use decimals. It’s considered because we have 10 fingers on our hands and this is perhaps true.

An interesting fact though, is that the Mayas (who barely predicted the end of the world a couple of weeks ago) used a system of a base 20. That is logical, since we have not 10, but total of 20 fingers considering our legs.

Finally, this algorithm may seem to easy to be explained but it lays down in more complex algorithms.

What’s the fastest way to sort the following sequence [9, 3, 0, 5, 4, 1, 2, 6, 8, 7]? Well, the question is a bit tricky since the input is somehow “predefined”. First of all we have only integers, and fortunately they are all different. That’s great and we know that in practice it’s almost impossible to count on such lucky coincidence. However here we can sort the sequence very quickly.

First of all we can pass through all these integers and by using an auxiliary array we can just put them at their corresponding index. We know in advance that that is going to work really well, because they are all different.

There is only one major problem in this solution. That’s because we assume all the integers are different. If not – we can just put all them in one single corresponding index.

That is why we can use bucket sort.

Overview

Bucket sort it’s the perfect sorting algorithm for the sequence above. We must know in advance that the integers are fairly well distributed over an interval (i, j). Then we can divide this interval in N equal sub-intervals (or buckets). We’ll put each number in its corresponding bucket. Finally for every bucket that contains more than one number we’ll use some linear sorting algorithm.

The thing is that we know that the integers are well distributed, thus we expect that there won’t be many buckets with more than one number inside.

That is why the sequence [1, 2, 3, 2, 1, 2, 3, 1] won’t be sorted faster than [4, 3, 1, 2, 9, 5, 4, 8].

Pseudo Code

1. Let n be the length of the input list L;
2. For each element i from L
   2.1. If B[i] is not empty
      2.1.1. Put A[i] into B[i] using insertion sort;
      2.1.2. Else B[i] := A[i] 
3. Concatenate B[i .. n] into one sorted list;

Complexity

The complexity of bucket sort isn’t constant depending on the input. However in the average case the complexity of the algorithm is O(n + k) where n is the length of the input sequence, while k is the number of buckets.

The problem is that its worst-case performance is O(n^2) which makes it as slow as bubble sort.

Application

As the other two linear time sorting algorithms (radix sort and counting sort) bucket sort depends so much on the input. The main thing we should be aware of is the way the input data is dispersed over an interval.

Another crucial thing is the number of buckets that can dramatically improve or worse the performance of the algorithm.

This makes bucket sort ideal in cases we know in advance that the input is well dispersed.

The first question when we see the phrase “sorting in linear time” should be – where’s the catch? Indeed there’s a catch and the thing is that we can’t sort just anything in linear time. Most of the time we can speak on sorting integers in linear time, but as we can see later this is not the only case.

Since we speak about integers, we can think of a faster sorting algorithm than usual. Such an algorithm is the counting sort, which can be very fast in some cases, but also very slow in others, so it can be used carefully. Another linear time sorting algorithm is radix sort.

Introduction

Count sort is absolutely brilliant and easy to implement. In case we sort integers in the range [n, m] on the first pass we just initialize a zero filled array with length m-n. Than on the second pass we “count” the occurrence of each integer. On the third pass we just sort the integers with an ease.

However we have some problems with that algorithm. What if we have only few items to sort that are very far from each other like [2, 1, 10000000, 2]. This will result in a very large unused data. So we need a dense integer sequence. This is important because we must know in advance the nature of the sequence which is rarely sure.

That’s why we need to use another linear time sorting algorithm for integers that doesn’t have this disadvantage. Such an algorithm is the radix sort.

Overview

The idea behind the radix sort is simple. We must look at our “integer” sequence as a string sequence. OK, to become clearer let me give you an example. Our sequence is [12, 2, 23, 33, 22]. First we take the leftmost digit of each number. Thus we must compare [_2, 2, _3, _3, _2]. Clearly we can assume that since the second number “2” is only a one digit number we can fill it up with a leading “0”, to become 02 or _2 in our example: [_2, _2, _3, _3, _2]. Now we sort this sequence with a stable sort algorithm.

What is a Stable Sort Algorithm

A stable sort algorithm is an algorithm that sorts a list by preserving the positions of the elements in case they are equal. In terms of PHP this means that:

array(0 => 12, 1=> 13, 2 => 12); 

Will be sorted as follows:

array(0 => 12, 2 => 12, 1 => 13);

Thus the third element becomes second following the first element. Note that the third and the first element are equal, but the third appears later in the sequence so it remains later in the sorted sequence.

In the radix sort example, we need a stable sort algorithm, because we need to worry about only one position of digit we explore.

So what happens in our example after we sort the sequence?

As we can see we’re far from a sorted sequence, but what if we proceed with the next “position” – the decimal digit?

Than we end up with this:

Now we have a sorted sequence, so let’s summarize the algorithm in a short pseudo code.

Pseudo Code

The simple approach behind the radix sort algorithm can be described as pseudo code, assuming that we’re sorting decimal integers.

1. For each digit at position 10^0 to 10^n
1.1. Sort the numbers by this digit using a stable sort algorithm;

The thing is that here we talk about decimal, but actually this algorithm can be applied equally on any numeric systems. That is why it’s called “radix” sort.

Thus we can sort binary numbers, hexadecimals etc.

It’s important to note that this algorithm can be also used to sort strings alphabetically.

[ABC, BBC, ABA, AC]
[__C, __C, __A, __C] => [ABA, ABC, BBC, AC]
[_B_, _B_, _B_, _A_] => [AC, ABA, ABC, BBC]
[___, A__, A__, B__] => [AC, ABA, ABC, BBC]

That is simply correct because we can assume that our alphabet is another 27 digit numeric system (in case of the Latin alphabet).

Complexity

As I said in the beginning radix sort is a linear time sorting algorithm. Let’s see why. First we depend on the numeric system. Let’s assume we have a decimal numeric system – then we have N passes sorting 10 digits which is simply 10*N. In case of K digit numeric system our algorithm will be O(K*N) which is linear.

However you must note that in case we sort N numbers in an N digit numeric system the complexity will become O(N^2)!

We must also remember that in order to implement radix sort and a supporting stable sort algorithm we need an extra space.

Application

Sorting integers can be faster than sorting just anything, so any time we need to implement a sorting algorithm we must carefully investigate the input data. And that’s also the big disadvantage of this algorithm – we must know the input in advance, which is rarely the case.

We already know what’s topological sort of a directed acyclic graph. So why do we need a revision of this algorithm? First of all I never mentioned its complexity, thus to understand why we do need a revision let’s get again on the algorithm.

We have a directed acyclic graph (DAG). There are no cycles so we must go for some kind of order putting all the vertices of the graph in such an order, that if there’s a directed edge (u, v), u must precede v in that order.

The process of putting all the vertices of the DAG in such an order is called topological sorting. It’s commonly used in task scheduling or while finding the shortest paths in a DAG.

The algorithm itself is pretty simple to understand and code. We must start from the vertex (vertices) that don’t have predecessors.

We put them in our sorted list in random order. Since they don’t depend on each other we can assume they are equally sorted already. Indeed thinking of a task schedule if there are tasks that don’t have predecessors (they don’t depend on other tasks before them) and that don’t depend on each other we can put them in random order (and execute them in random order).

Once we have the vertices with no predecessors we must remove the edges starting from them. Then – go again with the vertices with no predecessors.

It’s as simple as that, so why do we need a revision of this algorithm? Well, basically because of its efficiency.

Overview

As we know most of the graph algorithms depend on the way the graph is represented in our application. We consider as the two main representations the adjacency matrix …

… and adjacency lists.

Let’s first take a look of some of the main approaches to get the topologically sorted list at the end of the algorithm.

What can we do in order to find the vertices with no predecessors? We can only scan the entire list of vertices.

Adjacency Matrix

In case we’re using adjacency matrix we need|V|^2 space to store the graph. To find the vertices with no predecessors we have to scan the entire graph, which will cost us O(|V|^2) time. And we’ll have to do that |V| times. This will be |V|^3 time consuming algorithm and for dense graphs this will be quite an ineffective algorithm.

Adjacency Lists

What about the adjacency list? There we need |E| space to store a directed graph. How fast can we find a node with no predecessor? Practically we’ll need O(|E|) time. Thus in the worst case we have again O(|V|^2) time consuming programs.

So what can be done in order to optimize this algorithm?

Practically we can start by picking up a random vertex and “go back” until we get a node with no predecessors. This approach can be very effective yet also very ineffective. First of all if we have to scan all the way back to a node with no predecessors this will cost us |V| time, but if we stuck on a node that don’t have a preceding node then we’ll have a constant speed.

This means that we can modify the algorithm a bit in order to improve a lot the algorithm. We just need to store both incoming and outgoing edges and slightly modify the adjacency lists.

What’s the algorithm now?

First we easily find the nodes with no predecessors. Then, using a queue, we can keep the nodes with no predecessors and on each dequeue we can remove the edges from the node to all other nodes.

Pseudo Code

1. Represent the graph with two lists on each vertex (incoming edges and outgoing edges)
2. Make an empty queue Q;
3. Make an empty topologically sorted list T;
4. Push all items with no predecessors in Q;
5. While Q is not empty
   a. Dequeue from Q into u;
   b. Push u in T;
   c. Remove all outgoing edges from u;
6. Return T;

This approach will give us a better performance than the “brute force” approach. The running time complexity is O(|V| + |E|). The problem is that we need additional space and an operational queue, but this approach is a perfect example of how by using additional space you can get a better performing algorithm.

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.

The Strassen’s method of matrix multiplication is a typical divide and conquer algorithm. We’ve seen so far some divide and conquer algorithms like merge sort and the Karatsuba’s fast multiplication of large numbers. However let’s get again on what’s behind the divide and conquer approach.

Unlike the dynamic programming where we “expand” the solutions of sub-problems in order to get the final solution, here we are talking more on joining sub-solutions together. These solutions of some sub-problems of the general problem are equal and their merge is somehow well defined.

A typical example is the merge sort algorithm. In merge sort we have two sorted arrays and all we want is to get the array representing their union again sorted. Of course, the tricky part in merge sort is the merging itself. That’s because we’ve to pass through the two arrays, A and B, and we’ve to compare each “pair” of items representing an item from A and from B. A bit off topic, but this is the weak point of merge sort and although its worst-case time complexity is O(n.log(n)), quicksort is often preferred in practice because there’s no “merge”. Quicksort just concatenates the two sub-arrays. Note that in quicksort the sub-arrays aren’t with an equal length in general and although its worst-case time complexity is O(n^2) it often outperforms merge sort.

This simple example from the paragraph above shows us how sometimes merging the solutions of two sub-problems actually isn’t a trivial task to do. Thus we must be careful when applying any divide and conquer approach.

History

Volker Strassen is a German mathematician born in 1936. He is well known for his works on probability, but in the computer science and algorithms he’s mostly recognized because of his algorithm for matrix multiplication that’s still one of the main methods that outperforms the general matrix multiplication algorithm.

Strassen firstly published this algorithm in 1969 and proved that the n^3 algorithm isn’t the optimal one. Actually the given solution by Strassen is slightly better, but his contribution is enormous because this resulted in many more researches about matrix multiplication that led to some faster approaches, i.e. the Coppersmith-Winograd algorithm with O(n^2,3737).

Overview

The general algorithm on multiplying two matrices A[NxN] and B[NxN] is fairly simple. Although it’s more difficult than multiplying two numbers and also it is not commutative it’s still very simple – but slow.

Let’s first define what’s a matrix A[NxN]. As we speak about matrices NxN we usually think of a square grid with N rows and N columns. In each row and column A[i][j] we’ve a value.

Of course, as developers, we can think of a matrix as a two-dimensional array.

// PHP two-dimensional array
$a = array(
    0 => array($v1, $v2, $v3, $v4),
    1 => array($v5, $v6, $v7, $v8),
    2 => array($v9, $v10, $v11, $v12),
); 

Don’t forget that a NxN matrix is just a private case for a matrix. We can equally likely have any other size of a matrix NxM (N <> M).

However the size of a matrix is crucial in order to multiply it with another matrix. Why is that?

As I said above multiplying matrices isn’t the same as multiplying numbers. First of all this operation isn’t commutative.

And the second problem is the way you multiply two matrices A with B.

Just because this works with NxN matrices we can see the problem with multiplying rectangular matrices. Indeed, this wouldn’t be possible unless the second dimension of A isn’t exactly equal to the first dimension of B.

Hopefully we are now talking about square matrices with exactly the same dimensions.

OK, so now we know how to multiply two square matrices (with the same dimensions NxN) and now let’s evaluate the time complexity for the general purpose algorithm.

As we know A.B = C only when:

C[i][j] = sum(A[i][k] * B[k][j]) for k = 0 .. n

Thus we have n^3 operations. Let’s try to find out a divide and conquer approach.

Indeed this isn’t difficult in case of matrices because as we know we can divide in matrix in smaller sub-matrices.

Now what do we have?

Again – the same complexity – we have 8 products and 4 sums. Where’s the catch?

Of course in order to get faster solution we’ve to be looking as Strassen did in 1969. He defined P1, P2, P3, P4, P5, P6 and P7 as defined on the image below.

Complexity

As I mentioned above the Strassen’s algorithm is slightly faster than the general matrix multiplication algorithm. The general algorithm’s time complexity is O(n^3), while the Strassen’s algorithm is O(n^2.80).

You can see on the chart below how slightly faster is this even for large n.

Application

Although this algorithm seems to be more close to pure mathematics than to computer practically everywhere we use NxN arrays we can benefit from matrix multiplication.

In the other hand the algorithm of Strassen is not much faster than the general n^3 matrix multiplication algorithm. That’s very important because for small n (usually n < 45) the general algorithm is practically a better choice. However as you can see from the chart above for n > 100 the difference can be very big.

In the same time typically NxN arrays are used always when we talk about adjacency matrix of graphs |V| = n and some graph algorithms practically depend on matrix multiplication.

Along with the Kruskal’s minimum spanning tree algorithm, there’s another general algorithm that solves the problem. The algorithm of Prim.

As we already know the algorithm of Kruskal works in a pretty natural and logical way. Since we’re trying to build a MST, which is naturally build by the minimal edges of the graph (G), we sort them in a non-descending order and we start building the tree.

During the whole process of building the final minimum spanning tree Kruskal’s algorithm keeps a forest of trees. The number of trees in that forest decreases on each step and finally we get the minimum weight spanning tree.

A key point in the Kruskal’s approach is the way we get the “next” edge from G that should be added to one of the trees of the forest (or to connect two trees from the forest). The only thing we should be aware of is to choose an edge that’s connecting two vertices – u and v and these two shouldn’t be in the same tree. That’s all.

An important feature of the Kruskal’s algorithm is that it builds the MST just by sorting the edges by their weight and doesn’t care about a particular starting vertex.

In the same time there’s another algorithm that builds a MST – the algorithm of Prim designed by Robert Prim in 1957.

Overview

The idea behind the Prim’s algorithm is rather different from Kruskal’s approach. During the process of building the MST this algorithm keeps a single tree, which is finally sub-tree of the final minimum weight spanning tree.

On each step we chose an edge which we add to the growing tree that finally forms the MST.

It is somehow unnatural approach! We start from a given vertex and initially we don’t choose the lightest edge. Thus during the whole process the tree grows, but outside the tree (T) there might be edges that are lighter than those in the tree (i.e. the edge (5, 1) from the tree above is lighter than (2, 5) but (2, 5) is added to the growing tree before the edge (5, 1)).

Compared to the Kruskal’s algorithm this time everything seems to be really unnatural. How we should be sure the final tree (T) will be a minimum spanning tree since we don’t get the lightest edge on each step?

Actually we are sure that the final tree is a MST because of another obvious feature of the minimum spanning trees. They should “connect” all the vertices of G, thus somehow at least one edge reaching each vertex will appear in the MST. Thus we shouldn’t care where do we start, the only important thing is to choose the lightest edge that’s visible so far.

This algorithm looks much like Dijkstra’s shortest path in a graph, because we start from a vertex, we push all the edges starting from this node to a priority queue and we chose the lightest edge. Going to the next node connected by this edge we append to the queue all the edges that aren’t in the queue.

That way the queue grows and we get always the lightest edge – thus forming a priority queue.

Now let’s summarize the algorithm of Prim

Pseudo Code

As an initial input we have the graph (G) and a starting vertex (s).

1.  Make a queue (Q) with all the vertices of G (V);
2.  For each member of Q set the priority to INFINITY;
3.  Only for the starting vertex (s) set the priority to 0;
4.  The parent of (s) should be NULL;
5.  While Q isn’t empty
6.     Get the minimum from Q – let’s say (u); (priority queue);
7.     For each adjacent vertex to (v) to (u)
8.        If (v) is in Q and weight of (u, v) < priority of (v) then
9.           The parent of (v) is set to be (u)
10.          The priority of (v) is the weight of (u, v)

Indeed it looks much like the Dijkstra’s algorithm.

Code

Here’s a PHP implementation of the algorithm of Prim, which directly follows the pseudo code.

// Prim's algorithm

define('INFINITY', 100000000);

// the graph
$G = array(
    0 => array( 0,  4,  0,  0,  0,  0,  0,  0,  8),
    1 => array( 4,  0,  8,  0,  0,  0,  0,  0,  11),
    2 => array( 0,  8,  0,  7,  0,  4,  2,  0,  0),
    3 => array( 0,  0,  7,  0,  9,  14,  0,  0,  0),
    4 => array( 0,  0,  0,  9,  0,  10,  0,  0,  0),
    5 => array( 0,  0,  4,  14,  10,  0,  0,  2,  0),
    6 => array( 0,  0,  2,  0,  0,  0,  0,  6,  7),
    7 => array( 0,  0,  0,  0,  0,  2,  6,  0,  1),
    8 => array( 8,  11,  0,  0,  0,  0,  7,  1,  0),
);

function prim(&$graph, $start)
{
    $q = array(); // queue
    $p = array(); // parent
    
    foreach (array_keys($graph) as $k) {
        $q[$k] = INFINITY;
    }
    
    $q[$start] = 0;
    $p[$start] = NULL;
    
    asort($q);
    
    while ($q) {
        // get the minimum value
        $keys = array_keys($q);
        $u = $keys[0];
        
        foreach ($graph[$u] as $v => $weight) {
            if ($weight > 0 && in_array($v, $keys) && $weight < $q[$v]) {
                $p[$v] = $u;
                $q[$v] = $weight;
            }
        }
        
        unset($q[$u]);
        asort($q);
    }
    
    return $p;
}

prim($G, 5);

History

It’s curious to say that the algorithm developed by Robert Prim isn’t developed by him. It’s considered that a Czech mathematician Vojtech Jarnik discovered back in 1930. However now we know this algorithm as the algorithm of Prim, which independently discovered it in 1957 as I said above, and finally Edsger Dijkstra described it in 1959. That’s why his algorithm on finding the single-source shortest paths in a graph looks so much to this algorithm. Perhaps by finding this algorithm on minimum spanning tree Dijkstra discovered how we can find the shortest paths to all vertices using a priority queue. Indeed the paths to all other vertices use the edges of the minimum spanning tree.

Just because Jarnik found and described this algorithm 27 years earlier than Robert Prim, today it’s more convenient to call this algorithm the Prim-Jarnik algorithm.

One of the two main algorithms in finding the minimum spanning tree algorithms is the algorithm of Kruskal. Before getting into the details, let’s get back to the principles of the minimum spanning tree.

We have a weighted graph and of all spanning trees we’d like to find the one with minimal weight. As an example on the picture above you see a spanning tree (T) on the graph (G), but that isn’t the minimum weight spanning tree!

We can think of a group of islands and the possible connections of bridges connecting them. Of course building bridges is expensive and time consuming, so we must be aware of what kind of bridges we want to build. Nevertheless there is an important question, what’s the minimum price we’d like to pay to build such set of bridges connecting all the islands.

Thus we practically need to build a minimum spanning tree, where the vertices will be the islands, while the edges will be the possible bridges between them. Every possible bridge has a weight (the price or the time we need to build it, etc.).

This scenario is only one of possible use cases of where minimum spanning trees can be used in practice.

The two main approaches – the Kruskal’s and the Prim’s algorithms however differ.

Overview

The algorithm of Kruskal starts by initializing a set of |V| trees.

During the process of building the final spanning tree we keep a forest. Obviously we start with a forest with |V| trees, where each tree is a single node tree.

On some point we have a forest of “k” trees which are all a sub-trees of the minimum spanning tree.

Finally one step before building the final MST we have two trees and we connect them with the less weighted edge left that connects them.

It’s important to note that during the process of building the tree we sort the edges in ascending order by their weight.

Than we start getting edges and check whether their ends (the two vertices making the edge) belong to a different sub-trees.

Pseudo Code

1. T (the final spanning tree) is defined to be the empty set;
2. For each vertex v of G, make the empty set out of v;
3. Sort the edges of G in ascending (non-decreasing) order;
4. For each edge (u, v) from the sored list of step 3.
      If u and v belong to different sets
         Add (u,v) to T;
         Get together u and v in one single set;
5. Return T

A great feature about the Kruskal’s algorithm is that it also work on disconnected graphs.

History

Kruskal’s algorithm is named after Joseph Kruskal, who wasn’t only computer scientist, but also prominent mathematician and statistician. Although he is best known for its algorithm for computing the minimum spanning tree, described in this post, he’s also known with his work as a statistician and his contribution to the formulation of multidimensional scaling.

Kruskal also explored the Indo-European languages contributing the studies of the linguistics along with other scientists. His “Indo-European Lexicographical List” (http://www.wordgumbo.com/ie/cmp/) is still widely used.

Here’s a classical task on graphs. We have a group of cities and we must wire them to provide them all with electricity. Out of all possible connections we can make, which one is using minimum amount of wire.

To wire N cities, it’s clear that, you need to use at least N-1 wires connecting a pair of cities. The problem is that sometimes you have more than one choice to do it. Even for small number of cities there must be more than one solution as shown on the image bellow.

Here we can wire these four nodes in several ways, but the question is, which one is the best one. By the way defining the term “best one” is also tricky. Most often this means which uses least wire, but it can be anything else depending on the circumstances.

As we talk on weighted graphs we can generally speak of a minimum weight solution through all the vertices of the graph.

By the way there might be more the one equally optimal (minimal) solutions.

Overview

Obviously we must choose those edges that are enough to connect all the vertices of the graph and whose sum of weights is minimal. Since we can’t have cycles in our final solution it must form a tree. Thus we’re speaking on a minimum weight spanning tree, as the tree spans over the whole graph.

Does each connected and weighted graph have a minimum spanning tree? The answer is yes! By removing the cycles from the graph G we get a spanning tree, since it’s connected. From all possible spanning trees one or more are minimal.

If w(u, v) is the weight of the edge (u, v), we can speak of weight of any spanning tree T – w(T) which is the sum of all the edges forming that tree.

Thus the weight of the minimum spanning tree is less than the weight of whatever other spanning tree of G.

After we’re sure that there is at least one minimum spanning tree for all connected and weighted graphs we only need to find it somehow.

We can go with an incremental approach. At the end we’ll have the minimum spanning tree (MST), but before that on each step of our algorithm we’ll have a sub-set of this final tree, which will grow and grow until it becomes the real MST. This subset of edges we’ll keep in one additional set A.

So far we know that on each step we have a subset of the final MST, but first we need to answer a couple of questions.

How do we start?

Well, we’ll start with the empty set of edges. Clearly the empty set is a subset of any other set, thus it will be also a subset of the MST.

How do we grow the tree?

Another question we must answer is how to grow the tree. Since we have a MST sub-set (A) on each step how do we add an edge to this set in order to get another (bigger than the previous one) subset of edges, which will be again a subset of the minimum spanning tree?

Clearly we must make a decision which edge to add to the growing subset and this is the tricky part of this algorithm.

Chose the lowest weight edge!

To find the minimum spanning tree on each step we must get the lowest weighted edge that connects our subset (A) with the rest of the vertices.

However can we be sure that by choosing the less weighted edge we’ll get the MST? Well, let’s assume that isn’t right in order to prove that wrong!

OK so on some step of our growing sub-tree we don’t get the lightest edge (u, v), because we somehow doubt this rule, and we get another edge – let’s say (x, y). Mind that w(x, y) >= w(u, v).

Thus our final MST will contain somewhere in its set of edges the edge (x, y), but the weight of MST w(T) is minimal, and if we get another spanning tree that contains the exact same edges as T but instead of (x, y) contains (u, v) we’ll get a smaller weight!

That isn’t possible! Thus we proved that on each step we must get the less weighted edge.

This particular approach is called “greedy”, because on each step we get the best possible choice. However greedy algorithms don’t always get the right or optimal solution. Fortunately for MST this isn’t true so we can be greedy as much as we can!

OK let’s make a summary of our algorithm in the following pseudo code.

Pseudo Code

1. We start with an  empty set (A) subset of the final MST;
2. Until A does not form T:
      a. Get the less weighted edge u from G;	
      b. Add u to A;
3. Return A

Application

Actually this algorithm is used firstly by Borůvka which started to wire Moravia in 1926. Even without knowing that the “greedy” approach will lead him to the right solution he optimally covered Moravia with electricity.

However this algorithm is too general and there are two main algorithms – the Prim’s algorithm and the Kruskal’s algorithm that we shall see in future posts.

The thing is that on each step we must get the less weighted edge and both algorithms use different approaches to do that.