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.
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.
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.
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.
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!
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.