Question

Write a more detailed version of the dynamic programming algorithm for the Traveling Salesperson Problem (Algorithm 3.11 ).

Short answer

The dynamic programming solution to the Traveling Salesperson Problem involves setting up the states, decisions, and recursion suitably. The states are represented by the current city and the set of visited cities. The decision involves choosing the next city to visit. The recursion uses these to calculate the minimum cost. The dp table is filled iteratively in order of the size of the visited set for each city, and the final solution includes the cost to return to the first city.

Step-by-step solution

Set up the Problem

The Traveling Salesperson Problem involves finding the shortest possible route that allows a salesperson to visit a set of cities, each exactly once, and then return to the original city. Assume we have \(n\) cities, and let's label them as \(1, 2, ..., n\). Keep city 1 as the `home city`. The key variables are current city \(i\) and visited set \(S\).

Define the States

Define \(dp[i] [S]\) as the minimum cost to visit all cities in the set \(S\) exactly once, starting from city 1 and ending at city \(i\). At the start, only the home city has been visited, so \(S = \{1\}\). At the end, all cities have been visited, so \(S = \{1, 2, ..., n\}\).

Define the Decisions

For the decisions, at city \(i\) and visited set \(S\), the next city to visit is one that's not visited yet on the route. Assume city \(j\) is the next city to visit.

Define the Recursion

The recursion can be defined as \(dp[i][S] = min_{j \notin S} {dp[j][S \cup {j}]} + cost(i, j)\), that is the state \(dp[i][S]\) transition to another state \(dp[j][S\cup{j}]\) plus the cost to go from city \(i\) to city \(j\).

Define the Base Case

The base case would be for \(dp[1][\{1\}]\), which is set to 0, since the starting city is the home city and visiting the home city incurs no cost.

Define the Final Solution

The final solution, the minimum cost to visit all cities and return to the home city, is \(min_{i>1} {dp[i][\{1, 2, ..., n\}]} + cost(i, 1)\), to include the cost to return to the home city.

Iteratively Calculate the dp Table

Initially, fill in the dp table with large values. Then iteratively calculate, for each city \(i\) and each subset \(S\) of cities, \(dp[i][S]\) in increasing order of size. The order is crucial since the calculation of \(dp[i][S]\) depends on \(dp[j][S - \{j\}]\) for smaller subsets \(S - \{j\}\).