(a) Travelling Salesman Problem:
Given a set of cities and distance between every pair of cities, the problem is to find the shortest possible route or path that visits each city exactly once and returns to the starting point.
The problem is a famous NP hard problem in combinatorial optimization. Time complexity of travelling salesman is not polynomial but it is exponential. So, the problem is not solvable in polynomial time and hence, it comes under NP hard type problem.
Algorithm:
n=numbers of cities
m= n x n matrix of distances between cities
min=(infinity)
for all possible tours, do:
find the length of the tour
if length<min
min=length
store tour
Worst case (big O) time complexity of the given algorithm:
Solution to Travelling salesman problem:
Consider city 1 as the starting point and ending point.
Generate all (n-1)! Permutations of given cities.
Calculate cost of every permutation or tour and keep track of
minimum cost permutation.
Return the permutation or tour with minimum cost.
Let the given set of cities be {1, 2, 3, 4,….n}. Let us consider
city 1 as starting and ending point of output. For every other city
i (other than 1), we find the minimum cost path with city 1 as the
starting point, i as the ending point and all cities appearing
exactly once.
Let the cost of this route be cost(i), the cost of corresponding cycle would be cost(i) + dist(i,1) where dist(i, 1) is the distance from city i to 1. Finally, we got the minimum of all [cost(i) + dist(i, 1)] values.
To calculate cost (i) of route using Dynamic Programming, we need to have some recursive relation in terms of sub-problems. Let us define a term C(S, i) be the cost of the minimum cost route visiting each city in set S exactly once, starting at 1 and ending at i.
We start with all subsets of size 2 and calculate the cost C(S, i) for all subsets where S is the subset, then we calculate cost C(S, i) for all subsets S of size 3 and so on. Note that city 1 must be present in every subset. Calculate by following method:
If size of S is 2, then S must be {1, i},
C(S, i) = dist(1, i)
Else if size of S is greater than 2.
C(S, i) = min { cost C(S-{i}, j) + dis(j, i)} where j belongs to S, j != i and j != 1.
For a set of size n, we consider n-2 subsets each of size n-1 such that all subsets don’t have nth in them. So, total no. of subproblems are:
The Traveling Salesman problem (TSP) is famous. Given a list of cities and the distances in betwe...