# Travelling Salesman Algorithm

Brute Force Algorithm

A path through every vertex exactly once is the same as ordering the vertex in some way. Thus, to calculate the
minimum cost of travelling through every vertex exactly once, we can brute force every single one of the N!
permutations of the numbers from 1 to N.

```minimum = INF
for all permutations P
current = 0
for i from 0 to N-2
current = current + cost[P[i]][P[i+1]]   <- Add the cost of going from 1 vertex to the next

current = current + cost[P[N-1]][P[0]]           <- Add the cost of going from last vertex to the

first
if current < minimum            <- Update minimum if necessary
minimum = current

output minimum```

Time Complexity

There are N! permutations to go through and the cost of each path is calculated in O(N), thus this algorithm takes
O(N * N!) time to output the exact answer.

Dynamic Programming Algorithm

Notice that if we consider the path (in order): (1,2,3,4,6,0,5,7)
and the path (1,2,3,5,0,6,7,4)

The cost of going from vertex 1 to vertex 2 to vertex 3 remains the same, so why must it be recalculated? This result
can be saved for later use.
Let dp[ bitmask ] [ vertex ] represent the minimum cost of travelling through all the vertices whose corresponding
bit in bitmask is set to 1 ending at vertex. For example:

``````dp[12][2]

12 = 1 1 0 0
vertices: 3 2 1 0

Since 12 represents 1100 in binary, dp[12][2] represents going through vertices 2 and 3 in the graph with the path
ending at vertex 2.``````

Thus we can have the following algorithm (C++ implementation):

```int cost[N][N];                              //Adjust the value of N if needed
int memo[1 << N][N];                         //Set everything here to -1
{
int cost = INF;
if (bitmask == ((1 << N) - 1))           //All vertices have been explored
{
return cost[pos][0];                 //Cost to go back
}
{
return memo[bitmask][pos];	         //Just return the value, no need to recompute
}
for (int i = 0; i < N; ++i)              //For every vertex
{
if ((bitmask & (1 << i)) == 0)       //If the vertex has not been visited
{
cost = min(cost,TSP(bitmask | (1 << i) , i) + cost[pos][i]);         //Visit the vertex
}
}
memo[bitmask][pos] = cost;              //Save the result
return cost;
}

//Call TSP(1,0)```

This line may be a little confusing, so lets go through it slowly:
cost = min( cost, TSP( bitmask | (1 << i) , i ) + cost[ pos ] [ i ] );

Here, bitmask | (1 << i) sets the ith bit of bitmask to 1, which represents that the ith vertex has been visited. The
i after the comma represents the new pos in that function call, which represents the new “last” vertex. cost[ pos ][ i ] is to add the cost of travelling from vertex pos to vertex i.
Thus, this line is to update the value of cost to the minimum possible value of travelling to every other vertex that
has not been visited yet.

Time Complexity

The function TSP(bitmask,pos) has 2^N values for bitmask and N values for pos. Each function takes O(N) time to
run (the for loop). Thus this implementation takes O(N^2 * 2^N) time to output the exact answer.