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.