drilling.py

import networkx as nx

def get_distances(num_nodes):
    distances = {}
    for i in range(1, num_nodes+1):
        for j in range(1, num_nodes+1):
            distance = float(input(f"Enter the distance between node {i} and node {j}: "))
            distances[(i, j)] = distance
            distances[(j, i)] = distance
    return distances

def tsp_optimal_drilling(distances):
    G=nx.Graph()
    G.add_weighted_edges_from((i,j,distance) for (i,j),distance in distances.items())
    optimal_order = nx.approximation.traveling_salesman_problem(G,cycle=True)   
    return optimal_order

def calculate_optimal_cost(drill_order,distances):
    total_cost=sum(distances[(drill_order[i],drill_order[i+1])] for i in range(len(drill_order)-1))
    return total_cost

if __name__ == "__main__":
    while True:
        num_nodes = int(input("Enter th enumber of drill holes(nodes): "))
        distances = get_distances(num_nodes)
        optimal_order = tsp_optimal_drilling(distances)
        optimal_cost = calculate_optimal_cost(optimal_order,distances)
        print("Optimal order of drilling: ",optimal_order)
        print("Optimal cost of drilling: ",optimal_cost)
        try_again = input("Do you want to try again with different number of node? (yes/no):").lower()
        if try_again != "yes":
            break
         
"""
Enter the distance between node 1 and node 1: 2
Enter the distance between node 1 and node 2: 1
Enter the distance between node 1 and node 3: 3
Enter the distance between node 1 and node 4: 4
Enter the distance between node 2 and node 1: 5
Enter the distance between node 2 and node 2: 2
Enter the distance between node 2 and node 3: 1
Enter the distance between node 2 and node 4: 3
Enter the distance between node 3 and node 1: 2
Enter the distance between node 3 and node 2: 1
Enter the distance between node 3 and node 3: 2
Enter the distance between node 3 and node 4: 3
Enter the distance between node 4 and node 1: 6
Enter the distance between node 4 and node 2: 5
Enter the distance between node 4 and node 3: 1
Enter the distance between node 4 and node 4: 2
Optimal order of drilling:  [1, 3, 2, 3, 4, 3, 1]
Optimal cost of drilling:  8.0
"""