Consider the following 3 -PARTITION problem. Given integers...

Consider the following 3 -PARTITION problem. Given integers $a_{1}, \ldots, a_{n},$ we want to determine whether it is possible to partition of $\{1, \ldots, n\}$ into three disjoint subsets $I, J, K$ such that \[ \sum_{i \in I} a_{i}=\sum_{j \in J} a_{j}=\sum_{k \in K} a_{k}=\frac{1}{3} \sum_{i=1}^{n} a_{i} \] For example, for input (1,2,3,4,4,5,8) the answer is $y e s,$ because there is the partition (1,8) $(4,5),(2,3,4) .$ On the other hand, for input (2,2,3,5) the answer is no. Devise and analyze a dynamic programming algorithm for 3 -PARTITION that runs in time polynomial in $n$ and in $\sum_{i} a_{i}$

Consider the following 3 -PARTITION problem. Given integers $a_{1}, \ldots, a_{n},$ we want to determine whether it is possible to partition of $\{1, \ldots, n\}$ into three disjoint subsets $I, J, K$ such that
\[
\sum_{i \in I} a_{i}=\sum_{j \in J} a_{j}=\sum_{k \in K} a_{k}=\frac{1}{3} \sum_{i=1}^{n} a_{i}
\]
For example, for input (1,2,3,4,4,5,8) the answer is $y e s,$ because there is the partition (1,8) $(4,5),(2,3,4) .$ On the other hand, for input (2,2,3,5) the answer is no. Devise and analyze a dynamic programming algorithm for 3 -PARTITION that runs in time polynomial in $n$ and in $\sum_{i} a_{i}$

Key Concepts

Dynamic Programming

Dynamic programming is a problem-solving paradigm that involves breaking down a complex problem into smaller, overlapping subproblems, solving each subproblem once, and then storing their solutions – typically in a table – to avoid redundant computations. In the context of partition problems, dynamic programming is used to build up solutions for subsets with certain sum properties, thereby systematically exploring possible combinations that meet the partition criteria.

Subset Sum

The subset sum problem is a classic decision problem in which one determines whether a subset of numbers from a given set can sum up to a target value. This concept is crucial in partitioning problems, as finding subsets that add up to a particular fraction of the total sum underlies the strategy for dividing the set into multiple parts with equal sums.

Pseudo-Polynomial Time Algorithms

A pseudo-polynomial time algorithm is one whose running time is polynomial in the numeric value of the input (for example, the sum of the integers) rather than the size of the input in bits. In the dynamic programming approach to partition problems, the algorithm often depends on the sum of the integers, allowing it to be polynomial in both n and the total sum, even though the problem may be NP-complete in general.

State Space Representation

Designing a state space for dynamic programming involves defining the set of all relevant subproblem states and determining valid transitions between them. In partition problems, the state might encapsulate partial sums for different subsets, and transitions correspond to including new items in one of the subsets. This approach organizes the search for an equitable partition into manageable units that can be efficiently solved.

Transcript

00:01 Okay, so we want to express a brute force algorithm that determines if there is a disjoint pair in these subsets so i have this code right here and you can look at it, but i'll kind of describe this code holistically i'll describe what we're doing holistically and i'll describe the code as we go so first thing that we'll do is we first assume that there is no this this joint o -i -n pairs so we assume that we assume that there is no disjoints that there is no different pairs and as we run the code we want to prove otherwise.

00:36 So this whole code is proving that there is one destroyed pairs.

00:41 So the first thing that we do is we have s1, s2 and blah blah blah to sn.

00:48 So first thing that we'll do is we'll compare is s1 destroyed to s2, is s1 destroyed to s3 all the way down to sn.

00:57 And suppose that none of these are distraint.

01:01 So we then have to go to s2.

01:04 And we have to see if s2 is this joint with s3, s4, blah, blah, all the way to sn.

01:10 So now it's important that we don't actually compare to s1 because we've, yeah, we don't compare to s1 here because we've already compared s2 to s1 here already.

01:20 And you have to do the same thing for s4 compared to s5 and blah blah, blah, all the way to sn, then we compare sn minus 1 to sn.

01:28 This is our last one.

01:31 So that's what this first four loop does.

01:36 This line here, that's what this line does.

01:37 This first line loops through s1, s2, s4 to sn.

01:42 What is s3, sorry, s3 is s3, this is 4.

01:45 So that's from 1 to n minus 1...

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