Create Travelling Salesman Problem.py

Travelling Salesman Problem.py

@@ -0,0 +1,46 @@
+n = 4  # there are four nodes in example graph (graph is 1-based)
+
+# dist[i][j] represents shortest distance to go from i to j
+# this matrix can be calculated for any given graph using
+# all-pair shortest path algorithms
+dist = [[0, 0, 0, 0, 0], [0, 0, 10, 15, 20], [
+    0, 10, 0, 25, 25], [0, 15, 25, 0, 30], [0, 20, 25, 30, 0]]
+
+# memoization for top down recursion
+memo = [[-1]*(1 << (n+1)) for _ in range(n+1)]
+
+
+def fun(i, mask):
+    # base case
+    # if only ith bit and 1st bit is set in our mask,
+    # it implies we have visited all other nodes already
+    if mask == ((1 << i) | 3):
+        return dist[1][i]
+
+    # memoization
+    if memo[i][mask] != -1:
+        return memo[i][mask]
+
+    res = 10**9  # result of this sub-problem
+
+    # we have to travel all nodes j in mask and end the path at ith node
+    # so for every node j in mask, recursively calculate cost of
+    # travelling all nodes in mask
+    # except i and then travel back from node j to node i taking
+    # the shortest path take the minimum of all possible j nodes
+    for j in range(1, n+1):
+        if (mask & (1 << j)) != 0 and j != i and j != 1:
+            res = min(res, fun(j, mask & (~(1 << i))) + dist[j][i])
+    memo[i][mask] = res  # storing the minimum value
+    return res
+
+
+# Driver program to test above logic
+ans = 10**9
+for i in range(1, n+1):
+    # try to go from node 1 visiting all nodes in between to i
+    # then return from i taking the shortest route to 1
+    ans = min(ans, fun(i, (1 << (n+1))-1) + dist[i][1])
+
+print("The cost of most efficient tour = " + str(ans))
+