# Bitmasking and Dynamic Programming | Set-2 (TSP)

In this post, we will be using our knowledge of dynamic programming and Bitmasking technique to solve one of the famous NP-hard problem “Travelling Salesman Problem”.

Before solving the problem, we assume that the reader has the knowledge of

To understand this concept lets consider the below problem :

Problem Description
:

```Given a 2D grid of characters representing
a town where '*' represents the
houses, '#' represents the blockage,
'.' represents the vacant street
area. Currently you are (0, 0) position.

Our task is to determine the minimum distance
to be moved to visit all the houses and return
to our initial position at (0, 0). You can
only move to adjacent cells that share exactly
1 edge with the current cell.
```

The above problem is the well-known Travelling Salesman Problem.

The first part is to calculate the minimum distance between the two cells. We can do it by simply using a BFS as all the distances are unit distance. To optimize our solution we will be pre-calculating the distances taking the initial location and the location of the houses as the source point for our BFS.

Each BFS traversal takes O(size of grid) time. Therefore, it is O(X * size_of_grid) for overall pre-calculation, where X = number of houses + 1 (initial position)

Now lets’s think of a DP state

So we will be needing to track the visited houses and the last visited house to uniquely identify a state in this problem.

Therefore, we will be taking dp[index][mask] as our DP state.

Here,
index : tells us the location of current house
mask : tells us the houses that are visited ( if ith bit is set in mask then this means that the ith dirty tile is cleaned)

Whereas dp[index][mask] will tell us the minimum distance to visit X(number of set bits in mask) houses corresponding to their order of their occurrence in the mask where the last visited house is house at location index.

State transition relation

So our initial state will be dp[0][0] this tells that we are currently at initial tile that is our initial location and mask is 0 that states that no house is visited till now.

And our final destination state will be dp[any index][LIMIT_MASK], here LIMIT_MASK = (1<<N) – 1
and N = number of houses.

Therefore our DP state transition can be stated as

```dp(curr_idx)(curr_mask) = min{
}
```

The above relation can be visualized as the minimum distance to visit all the houses by standing at curr_idx house and by already visiting cur_mask houses is equal to min of distance between the curr_idx house and idx house + minimum distance to visit all the houses by standing at idx house and by already
visiting ( cur_mask | (1 <<idx) ) houses.

So, here we iterate over all possible idx values such that cur_mask has ith bit as 0 that tells us that ith house is not visited.

Whenever we have our mask = LIMIT_MASK, this means that we have visited all the houses in the town. So, we will add the distance from the last visited town (i.e the town at cur_idx positon) to the initial position (0, 0).

The C++ program for the above implementation is given below:

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

```#include <bits/stdc++.h>
using namespace std;

#define INF 99999999
#define MAXR 12
#define MAXC 12
#define MAXHOUSE 12

// stores distance taking souce
// as every dirty tile
int dist[MAXR][MAXC][MAXHOUSE];

// memoization for dp states

// stores coordinates for
// dirty tiles
vector < pair < int, int > > dirty;

// Directions
int X[] = {-1, 0, 0, 1};
int Y[] = {0, 1, -1, 0};

char arr[21][21];

// len : number of dirty tiles + 1
// limit : 2 ^ len -1
// r, c : number of rows and columns
int len, limit, r, c;

// Returns true if current position
// is safe to visit
// else returns false
// Time Complexity : O(1)
bool safe(int x, int y)
{
if (x >= r or y>= c or x<0 or y<0)
return false;
if (arr[x][y] == '#')
return false;
return true;
}

// runs BFS traversal at tile idx
// calulates distance to every cell
// in the grid
// Time Complexity : O(r*c)
void getDist(int idx){

// visited array to track visited cells
bool vis[21][21];
memset(vis, false, sizeof(vis));

// getting current positon
int cx = dirty[idx].first;
int cy = dirty[idx].second;

// initializing queue for bfs
queue < pair < int, int > > pq;
pq.push({cx, cy});

// initializing the dist to max
// because some cells cannot be visited
// by taking source cell as idx
for (int i = 0;i<= r;i++)
for (int j = 0;j<= c;j++)
dist[i][j][idx] = INF;

// base conditions
vis[cx][cy] = true;
dist[cx][cy][idx] = 0;

while (! pq.empty())
{
auto x = pq.front();
pq.pop();
for (int i = 0;i<4;i++)
{
cx = x.first + X[i];
cy = x.second + Y[i];
if (safe(cx, cy))
{
if (vis[cx][cy])
continue;
vis[cx][cy] = true;
dist[cx][cy][idx] = dist[x.first][x.second][idx] + 1;
pq.push({cx, cy});
}
}
}
}

// Dynamic Programming state transition recursion
// with memoization. Time Complexity: O(n*n*2 ^ n)
{
// goal state
return dist[0][0][idx];

int ret = INT_MAX;

// state transiton relation
for (int i = 0;i<len;i++)
{
if ((mask & (1<<i)) == 0)
{
ret = min( ret, solve(i, newMask)
+ dist[dirty[i].first][dirty[i].second][idx]);
}
}

}

void init()
{
// initializing containers
memset(dp, -1, sizeof(dp));
dirty.clear();

// populating dirty tile positions
for (int i = 0;i<r;i++)
for (int j = 0;j<c;j++)
{
if (arr[i][j] == '*')
dirty.push_back({i, j});
}

// inserting ronot's location at the
// begining of the dirty tile
dirty.insert(dirty.begin(), {0, 0});

len = dirty.size();

limit = (1<<len) - 1;

// precalculating distances from all
// dirty tiles to each cell in the grid
for (int i = 0;i<len;i++)
getDist(i);
}

int main(int argc, char const *argv[])
{
// Test case #1:
//     .....*.
//     ...#...
//     .*.#.*.
//     .......

char A[4][7] = {    {'.', '.', '.', '.', '.', '*', '.'},
{'.', '.', '.', '#', '.', '.', '.'},
{'.', '*', '.', '#', '.', '*', '.'},
{'.', '.', '.', '.', '.', '.', '.'}
};

r = 4; c = 7;

cout << "The given grid : " << endl;

for (int i = 0;i<r;i++)
{
for (int j = 0;j<c;j++)
{
cout << A[i][j] << " ";
arr[i][j] = A[i][j];
}
cout << endl;
}

// - initializitiation
// - precalculations
init();

int ans = solve(0, 1);

cout << "Minimum distance for the given grid : ";
cout << ans << endl;

// Test Case #2
//     ...#...
//     ...#.*.
//     ...#...
//     .*.#.*.
//     ...#...

char Arr[5][7] = {  {'.', '.', '.', '#', '.', '.', '.'},
{'.', '.', '.', '#', '.', '*', '.'},
{'.', '.', '.', '#', '.', '.', '.'},
{'.', '*', '.', '#', '.', '*', '.'},
{'.', '.', '.', '#', '.', '.', '.'}
};

r = 5; c = 7;

cout << "The given grid : " << endl;

for (int i = 0;i<r;i++)
{
for (int j = 0;j<c;j++)
{
cout << Arr[i][j] << " ";
arr[i][j] = Arr[i][j];
}
cout << endl;
}

// - initializitiation
// - precalculations
init();
ans = solve(0, 1);
cout << "Minimum distance for the given grid : ";
if (ans >= INF)
cout << "not possible" << endl;
else
cout << ans << endl;

return 0;
}
```

Output:

```The given grid :
. . . . . * .
. . . # . . .
. * . # . * .
. . . . . . .
Minimum distance for the given grid : 16
The given grid :
. . . # . . .
. . . # . * .
. . . # . . .
. * . # . * .
. . . # . . .
Minimum distance for the given grid : not possible
```

Note:

We have used the initial state to be dp[0][1] because we have pushed the start location at the first position in the container of houses. Hence, our Bit Mask will be 1 as the 0th bit is set i.e we have visited the starting location for our trip.

Time Complexity:

Consider the number of houses to be n. So, there are n * (2n) states and at every state, we are looping over n houses to transit over to next state and because of memoization we are doing this looping transition only once for each state. Therefore, our Time Complexity is O(n2 * 2n).

Recommended:

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