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 G {v₁, .... v_n}, we have two situations:

1. v_n belongs to the solution. In this case, v_n-1 does not belong to the solution, by the properties of an independent set. The maximum weight of the independent set for the graph G is W = w_n + W^'', where W^'' is the maximum weight independent set for the n-2 vertices path graph {v₁, .... v_n-2}.

2. v_n does not belong to the solution, and in this case the solution consists in the maximum weight independent set of the graph {v₁, .... v_n-1}, annotated as W^'.

There is no other possibility, so if we compute recursively W^' and W^'' we can decide, after the computation is completed, 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.

Dynamic Programming Algorithm

The actual solution involves computing the maximum weights W of the independent set starting with the vertex v₁ and storing the intermediate solutions in the array W:

initialize W[n]
W[0] = 0
W[1] = w1
for i = 2 to n:
  W[i] = max(W[i-1], w[i] + W[i-2])