This is my first post of 2019 and my first post in a while. There was one posted a few months ago, but not really geared towards the algorithms and learning focus of the site. I have been doing a lot of coding in my spare time, but honestly life has just gotten in the way. Its not a bad thing, but life is life and sometimes I have to prioritize the things. In particular, I have been having ideas and actually coding things up but the time it takes to clean up code, write a blog entry and finding a nice way to visualize these things has been something that I haven’t been able to really focus on as much as I’ve wanted to.

That said, I want to talk to you about the Set Partition Problem today. You can go to Wikipedia to get more information about this problem, but I will give you a brief introduction to it and then talk about two different approaches to it. The problem assumes that we are given as input a (multi)-set S. The reason we say it is a multi-set and not a simple set is because we can have the same element appear multiple times in the set. So if S_{1} = {2} and S_{2} = {2, 2}, then although as sets they are both equal to the set {2} = S_{1}, as multi-sets allow for multiple instances of an element. The elements of S are assumed to be positive integers.

So given this multi-set S, we ask the question of can the elements of S be divided into two smaller multi-sets, C_{1} and C_{2} where

- C
_{1}[union] C_{2}= S - C
_{1}[intersect] C_{2}= [empty set] - [Sigma]_[x in C
_{1}] = [Sigma]_[x in C_{2}].C_{1}[union] C_{2}= S, C_{1}[intersect] C_{2}= [empty set], and [Sigma]_[x in C_{1}] = [Sigma]_[x in C_{2}]

The first two bullets above say that the sets C_{1} and C_{2} form a partition of S. The third bullet says that the sums of the elements in the two children multi-sets are equal.

This problem is known to be NP Complete. This means that it is one of the more difficult decision problems. Because of this finding an algorithm that solves this problem exactly will generally take a long running time. And finding an algorithm that runs quickly will more than likely be incorrect in some instances.

I will show you two approaches to this problem. One is based on Dynamic Programming (DP) and one is based on Greedy Algorithms. The DP version solves the problem exactly but has a slow running time. The greedy algorithm is fast but is not guaranteed to always give the correct answer.

**Dynamic Programming **is based on the principle of optimality. This says that in order to have a correct solution to the overall problem, we need optimal solutions to all subproblems of this problem. This is done by keeping track of a table which can be used for looking up these subproblems and the optimal solutions to these subproblems.

For this problem, we will build a table where the rows represent the final sums and columns represent subsets of the given multi-set (containing the first 0…n elements). The question we are repeatedly asking is “can we find a subset of the set in this column whose sum is exactly the given rowsum?” Here is an example of the DP algorithm on the multi-set {5, 6, 5, 6, 7}.

**Greedy Algorithms **are generally based on sorting the elements based on some principle and using that to try to answer the underlying question. The main problem with this approach is that it is very short sided because they do not look at the overall picture. Such algorithms are known for finding local optima that are not always globally optimal. The benefit to these problems though is that they are generally easier to code and easier to understand.

For the Set Partition Problem , the Greedy approach is to sort elements in descending order. Once this is done, the goal is to keep two subsets while iterating through the array, adding the element to the smaller of the two sets whenever possible.

For more examples, check out Set Partition Problems