Programming Assignment 6 from Dr. H's 202 class
The problem here is described as follows: Given a list of cities and the cost to traverse between each city, where one can traverse from any city to any other city, find the most cost efficient route to visit each city once. This problem is commonly known as the traveling salesman problem. In this specific assignment, the aim is to use commonly known as a Dijkstra Approximation. An algorithm that finds the shortest path between nodes (cities in this case) by traveling to the closest node one is currently at each step. In effect, it ignores the overall cost of the journey and only takes into account the current leg of the trip. It can, and often does, miss the absolute most cost-efficient path through a graph; however, it is a brilliantly simple method and gets very close to the true shortest path. As oppesed to a depth-first-search ( O(N!) ), this algorithm has a time complexity of O(N^2) if each element's closest city is at the end of its row in the adjacency matrix (worst case).
In this implementation, the data for the cost/distance between nodes is represented in a two dimensional adjacency matrix. Value[x, y] on the matrix represents the distance from x -> y. The program retireves the information from text files and populates the matrix once at runtime and that is the only time the entire matrix is addressed iteratively. The actual cost algorithm utilizes a stack data structure that allows for efficient retrival as only the top element is needed (along with a couple helper data containers to keep track of which city has been visited). Once all cities have been visited, the program clears the stack and returns the cost of its trip.
This approach is the epitimy of what a dynamic programming model can do to a problem. The traveling salesman problem is effectively unsolvable as in order to exhaust all solutions for large N, one must have computing power that is simply not possible. This approach, while it doesn't guarentee the absolute best soultion, gives an approximation that may as well be.
12 city matrix
0 -> 5 -> 3 -> 8 -> 4 -> 1 -> 11 -> 6 -> 7 -> 10 -> 9 -> 2
Cost: 715
13 city matrix
0 -> 5 -> 3 -> 8 -> 4 -> 1 -> 11 -> 6 -> 7 -> 10 -> 9 -> 2 -> 12
Cost: 804
14 city matrix
0 -> 5 -> 3 -> 8 -> 4 -> 1 -> 13 -> 11 -> 6 -> 7 -> 10 -> 9 -> 2 -> 12
Cost: 900
15 city matrix
0 -> 5 -> 3 -> 8 -> 4 -> 1 -> 13 -> 14 -> 12 -> 2 -> 9 -> 10 -> 7 -> 6 -> 11
Cost: 840
16 city matrix
0 -> 5 -> 11 -> 8 -> 4 -> 1 -> 9 -> 3 -> 14 -> 13 -> 10 -> 15 -> 12 -> 7 -> 6 -> 2
Cost: 1418
19 city matrix
0 -> 5 -> 11 -> 8 -> 4 -> 1 -> 9 -> 3 -> 14 -> 18 -> 15 -> 12 -> 7 -> 6 -> 10 -> 13 -> 17 -> 16 -> 2
Cost: 1455
29 city matrix
0 -> 27 -> 5 -> 11 -> 8 -> 4 -> 20 -> 1 -> 19 -> 9 -> 3 -> 14 -> 18 -> 24 -> 6 -> 22 -> 26 -> 23 -> 7 -> 15 -> 12 -> 17 -> 13 -> 21 -> 16 -> 10 -> 28 -> 25 -> 2
Cost: 1800