The Traveling Salesman Problem (TSP) is one of the most classic and challenging problems in computing. We will solve it using a dynamic programming approach with bit masking. This method is efficient for a moderate number of cities, usually up to 20 or 25 cities. ## <br>Problem description Given a set of 𝑛 n cities and a distance matrix dist dist where dist [ 𝑖 ] [ 𝑗 ] dist[i][j] represents the distance between the city 𝑖 i and the city 𝑗 j, the goal is to find the shortest route that visits all the cities exactly once and returns to the city of origin. ## <br>Approach We will use the Bit Mask Dynamic Programming technique. The main idea is to use a bit mask to represent the set of cities visited so far. The state function will be defined as: dp [ 𝑚 𝑎 𝑠 𝑘 ] [ 𝑖 ] = lowest cost to visit all cities in the set 𝑚 𝑎 𝑠 𝑘 ending in the city 𝑖 dp[mask][i]=lowest cost to visit all cities in the mask set ending in city i The transition will be made by trying to visit a new city 𝑗 j from city 𝑖 i. ## <br>Implementation steps 1. **Initialization**: Initialize the dp matrix with infinite values and dp[1][0] = 0, meaning that we start at city 0 with only city 0 visited. 2. **Transition**: For each mask and each city 𝑖 i already visited, try to visit a new city 𝑗 j that has not yet been visited and update the cost. 3. **Final answer**: The answer will be the lowest cost to visit all the cities and return to the city of origin. ## <br>Python implementation Here is a Python implementation to solve the TSP using the technique described: ```py def tsp(dist): n = len(dist) # Initialize dp array dp = [[float('inf')] * n for _ in range(1 << n)] dp[1][0] = 0 # Start from city 0 # Iterate over all possible masks for mask in range(1 << n): for i in range(n): if mask & (1 << i): for j in range(n): if not mask & (1 << j): next_mask = mask | (1 << j) dp[next_mask][j] = min(dp[next_mask][j], dp[mask][i] + dist[i][j]) # Find the minimum cost to return to the starting city final_mask = (1 << n) - 1 return min(dp[final_mask][i] + dist[i][0] for i in range(1, n)) # Example usage dist = [ [0, 10, 15, 20], [10, 0, 35, 25], [15, 35, 0, 30], [20, 25, 30, 0] ] print(f"The lowest cost of the Traveling Salesman path is: {tsp(dist)}") ``` ## <br>Implementation explained 1. **Initialization**: **dp** is a two-dimensional matrix where **dp[mask][i]** represents the lowest cost to visit all the cities in the **mask** set ending in city **i**. 2. **Iteration over Masks**: We iterate over all possible masks (subsets of cities). 3. **Updating States**: For each city **i** in the **mask** set, we try to visit a new city **j** that is not in the **mask** set. We update **dp[next_mask][j]** if we find a shorter route. 4. **Final answer**: The final answer is the lowest cost to visit all the cities and return to the city of origin. This approach is efficient for up to approximately 20 cities due to the exponential growth of the state space, but it is an exact solution to the TSP problem. I would like to see your corrections if there is a need to do so or your comments on the solution to the problem