Fixes #12128

graphs/travlelling_salesman_problem.py

@@ -0,0 +1,155 @@
+"""
+title : Travelling Sales man Problem
+references : https://en.wikipedia.org/wiki/Travelling_salesman_problem
+author : SunayBhoyar
+"""
+
+import itertools
+import math
+
+demo_graph_points = {
+    "A": [10, 20],
+    "B": [30, 21],
+    "C": [15, 35],
+    "D": [40, 10],
+    "E": [25, 5],
+    "F": [5, 15],
+    "G": [50, 25],
+}
+
+
+# Euclidean distance - shortest distance between 2 points
+def distance(point1, point2):
+    """
+    Calculate the Euclidean distance between two points.
+    @input: point1, point2 (coordinates of two points as lists [x, y])
+    @return: Euclidean distance between point1 and point2
+    @example:
+    >>> distance([0, 0], [3, 4])
+    5.0
+    """
+    return math.sqrt((point2[0] - point1[0]) ** 2 + (point2[1] - point1[1]) ** 2)
+
+
+"""
+Brute force
+Time Complexity - O(n!×n)
+Space Complexity - O(n)
+"""
+
+
+def travelling_sales_man_problem_brute_force(graph_points):
+    """
+    Solve the Travelling Salesman Problem using brute force (permutations).
+    @input: graph_points (dictionary with node names as keys and coordinates as values)
+    @return: shortest path, total distance (list of nodes representing the shortest path and the distance of that path)
+    @example:
+    >>> travelling_sales_man_problem_brute_force({'A': [0, 0], 'B': [0, 1], 'C': [1, 0]})
+    (['A', 'B', 'C', 'A'], 3.414)
+    """
+    nodes = list(graph_points.keys())
+
+    min_path = None
+    min_distance = float("inf")
+
+    # Considering the first Node as the start position
+    start_node = nodes[0]
+    other_nodes = nodes[1:]
+
+    for perm in itertools.permutations(other_nodes):
+        path = [start_node] + list(perm) + [start_node]
+        total_distance = 0
+        for i in range(len(path) - 1):
+            total_distance += distance(graph_points[path[i]], graph_points[path[i + 1]])
+
+        if total_distance < min_distance:
+            min_distance = total_distance
+            min_path = path
+
+    return min_path, min_distance
+
+
+"""
+dynamic_programming
+Time Complexity - O(n^2×2^n)
+Space Complexity - O(n×2^n)
+"""
+
+
+def travelling_sales_man_problem_dp(graph_points):
+    """
+    Solve the Travelling Salesman Problem using dynamic programming.
+    @input: graph_points (dictionary with node names as keys and coordinates as values)
+    @return: shortest path, total distance (list of nodes representing the shortest path and the distance of that path)
+    @example:
+    >>> travelling_sales_man_problem_dp({'A': [0, 0], 'B': [0, 1], 'C': [1, 0]})
+    (['A', 'B', 'C', 'A'], 3.414)
+    """
+    n = len(graph_points)
+    nodes = list(graph_points.keys())
+
+    # Precompute distances between every pair of nodes
+    dist = [[0] * n for _ in range(n)]
+    for i in range(n):
+        for j in range(n):
+            dist[i][j] = distance(graph_points[nodes[i]], graph_points[nodes[j]])
+
+    # dp[mask][i] represents the minimum distance to visit all nodes in the 'mask' set, ending at node i
+    dp = [[float("inf")] * n for _ in range(1 << n)]
+    dp[1][0] = 0  # Start at node 0
+
+    # Iterate over all subsets of nodes (represented by mask)
+    for mask in range(1 << n):
+        for u in range(n):
+            if mask & (1 << u):
+                for v in range(n):
+                    if mask & (1 << v) == 0:
+                        next_mask = mask | (1 << v)
+                        dp[next_mask][v] = min(
+                            dp[next_mask][v], dp[mask][u] + dist[u][v]
+                        )
+
+    # Reconstruct the path and find the minimum distance to return to the start
+    final_mask = (1 << n) - 1
+    min_cost = float("inf")
+    end_node = -1
+
+    # Find the minimum distance from any node back to the starting node
+    for u in range(1, n):
+        if min_cost > dp[final_mask][u] + dist[u][0]:
+            min_cost = dp[final_mask][u] + dist[u][0]
+            end_node = u
+
+    # Reconstruct the path using the dp table
+    path = []
+    mask = final_mask
+    while end_node != 0:
+        path.append(nodes[end_node])
+        for u in range(n):
+            if (
+                mask & (1 << u)
+                and dp[mask][end_node]
+                == dp[mask ^ (1 << end_node)][u] + dist[u][end_node]
+            ):
+                mask ^= 1 << end_node
+                end_node = u
+                break
+
+    path.append(nodes[0])
+    path.reverse()
+
+    return path, min_cost
+
+
+if __name__ == "__main__":
+    print(f"Travelling salesman problem solved using Brute Force:")
+    path, distance_travelled = travelling_sales_man_problem_brute_force(
+        demo_graph_points
+    )
+    print(f"Shortest path: {path}")
+    print(f"Total distance: {distance_travelled:.2f}")
+
+    print(f"\nTravelling salesman problem solved using Dynamic Programming:")
+    path, distance_travelled = travelling_sales_man_problem_dp(demo_graph_points)
+    print(f"Shortest path: {path}")
+    print(f"Total distance: {distance_travelled:.2f}")