Maximum Weight Independent Set Problem

External

• https://www.coursera.org/learn/algorithms-greedy/lecture/t9XAF/wis-in-path-graphs-optimal-substructure
• https://www.coursera.org/learn/algorithms-greedy/lecture/w040v/wis-in-path-graphs-a-linear-time-algorithm
• https://www.coursera.org/learn/algorithms-greedy/lecture/TZgJM/wis-in-path-graphs-a-reconstruction-algorithm

Internal

• Dynamic Programming Algorithms

Overview

This article introduces the maximum weight independent set of a path graph and provides a dynamic programming algorithm to solve it.

The Maximum Weight Independent Set Problem

Given a path graph G=(V, E) where V consists in a set of n vertices v₀, v₁ ... v_n-1 that form a path, each of vertices with its own positive weight w_i, compute a maximum weight independent set of the graph. An independent set is a set of vertices in which none is adjacent to the other.

A Dynamic Programming Approach

The key to finding a dynamic programming algorithm is to identify a small set of subproblems whose solution can be computed using the previous subproblems' solutions.

In this case, we start with the observation that for the full n vertex path graph {v₀, .... v_n-1}, we have two situations:

1. v_n-1 belongs to the solution. In this case, v_n-2 does not belong to the solution, by the properties of an independent set, and the maximum weight of the independent set for the graph G_n is MXW_n = w_n-1 + MXW_n-2, where MXW_n-2 is the maximum weight independent set for the path graph {v₀, .... v_n-3}.

2. v_n-1 does not belong to the solution, and in this case the solution consists in the maximum weight independent set of the graph {v₀, .... v_n-2}, which is MXW_n-1.

There is no other possibility, so if we compute recursively MXW_n-2 and MXW_n-1 we can decide at this which one is larger.

This idea might suggest a recursive solution of the algorithm, but solution is inefficient, the running time is exponential. The inefficiency of the solution comes from the fact that both recurrences compute redundantly almost the same thing.