Question

I have read from multiple sources and from my understanding of the algorithm that it runs in 2^N time. My question is what causes TSP to achieve this run time? I can’t seem to find a pseudo-code so i can examine it.

Answer 1

The algorithm you mean is likely the inclusion-exclusion:

Find the shortest path though the following state space using A*:

• the state is defined by the partition of the set of cities to the ‘visited’ set, ‘unvisited’ set and the ‘current’ node.
• a valid transition is one that moves one node from ‘current’ to ‘visited’ and one from the ‘unvisited’ to the ‘current’ set. Its cost is equal the distance from the old ‘current’ to the new ‘current’.
• the starting state is: no city is ‘visited’, an arbitrary city is ‘current’.
• a finishing state is: no city is ‘unvisited’, any city is ‘current’.

The time complexity of inclusion-exclusion is given by the number of states: there is exactly one ‘current’ city (factor of n) and all other cities are either visited or unvisited (factor of 2^n).

The ‘A*’ algorithm will enter each state at most once. For each state, it will explore at most ‘n’ other nodes and push them into the priority queue. The priority queue will take at most ‘O(n)’ time to perform its operation.

Thus, the running time is O(2^n * n * n * O(n)) = O(2^n * poly(n)). Further insight shows that O(2^n * poly(n)) is equal to O(2^n).