C# Coding Test Course, Salesman's Dilemma

Hello, everyone! In this post, we will discuss a coding test problem using C#. The topic is “The Salesman’s Dilemma.” This problem is a common path optimization problem in algorithms, which is useful for understanding the important concepts of optimal path search and performance improvement.

Problem Description

A salesman needs to visit N cities. The distance information between each city is given, and the salesman must return to the starting city after visiting all cities once. The salesman needs to find a path that visits all cities with the minimum distance. Thus, this problem corresponds to the ‘Travelling Salesman Problem (TSP)’ based on the given cities and distance information.

Problem Definition

Given Input:
1. N (1 <= N <= 10, number of cities)
2. A 2D array dist[n][n] representing the distance information between each city (0 <= dist[i][j] <= 10^4)

Output:
1. The minimum distance value that passes through all cities

Example Input


N = 4
dist = [[0, 10, 15, 20],
         [10, 0, 35, 25],
         [15, 35, 0, 30],
         [20, 25, 30, 0]]

Explanation of this example:

In this array, dist[i][j] represents the distance from city i to city j. The salesman must visit all cities and return with the minimum distance. For example, there are various paths that can go from the starting city to city 1, then to city 2, and finally to city 3 before returning. Our goal is to find and output the shortest path among those options.

Approach to Solve the Problem

This problem can be solved using a backtracking approach with DFS (Depth-First Search). However, since N is limited to 10 or less, it is sufficient to use a brute-force technique. The most important part of this problem is determining which path to take, which depends on how the given distance array is handled.

Algorithm Steps

Keep track of the number of cities.
Explore all possible scenarios of the remaining cities based on the current city.
Update the path with the least distance.
Finally, return the minimum distance.

C# Code Example

using System;

class Salesman
{
    static int N;
    static int[,] dist;
    static bool[] visited;
    static int minCost = int.MaxValue;

    static void Main(string[] args)
    {
        N = 4;
        dist = new int[,] {
            { 0, 10, 15, 20 },
            { 10, 0, 35, 25 },
            { 15, 35, 0, 30 },
            { 20, 25, 30, 0 }
        };

        visited = new bool[N];
        visited[0] = true; // The starting city has been visited.
        DFS(0, 0, 1); // Visit the first city
        Console.WriteLine("Minimum distance: " + minCost);
    }

    static void DFS(int currentCity, int cost, int count)
    {
        // If all cities have been visited
        if (count == N)
        {
            // Calculate the return distance
            cost += dist[currentCity, 0];
            minCost = Math.Min(minCost, cost);
            return;
        }

        // Explore the next city
        for (int nextCity = 0; nextCity < N; nextCity++)
        {
            if (!visited[nextCity])
            {
                visited[nextCity] = true; // Mark the city as visited
                DFS(nextCity, cost + dist[currentCity, nextCity], count + 1);
                visited[nextCity] = false; // Backtracking
            }
        }
    }
}

Code Explanation

The code above is an example that uses the DFS backtracking technique to find the minimum distance:

Main method: Initializes the number of cities and the distance array, then starts the DFS from the first city.
DFS method: Recursively explores unvisited cities from the current city. If all cities have been visited, it adds the return trip to the start city and updates the minimum cost.

Complexity Analysis

The time complexity of this algorithm is O(N!). This is because it explores all possible paths. However, since N is small, it is feasible to execute.

Test Cases

Let’s verify this algorithm with various test cases:

// Test case 1
N = 3
dist = [[0, 10, 15],
         [10, 0, 20],
         [15, 20, 0]]
// Expected output: 35

// Test case 2
N = 5
dist = [[0, 20, 30, 10, 50],
         [20, 0, 40, 30, 50],
         [30, 40, 0, 30, 20],
         [10, 30, 30, 0, 20],
         [50, 50, 20, 20, 0]]
// Expected output: 90

Conclusion

Through this coding test problem solution, we have resolved the salesman’s dilemma. By understanding the algorithm and implementing it in C#, we learned how to utilize the distance array effectively. The salesman problem is often featured in algorithm competitions, so be sure to practice enough. Additionally, trying to solve similar problems using various approaches will also be good practice.

If you found this post useful, please look forward to the next one! Thank you.

C# Coding Test Course, Salesman’s Dilemma