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Bitmasking and Dynamic Programming | Travelling Salesman Problem
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Bitmasking and Dynamic Programming | Travelling Salesman Problem

Last Updated : 26 Nov, 2024
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Given a 2D grid of characters representing a town where '*' represents the houses, 'x' represents the blockage, '.' represents the vacant street area and 'o' represents the starting position. The task is to determine the minimum distance to be moved to visit all the houses if possible otherwise return -1. We can only move to adjacent cells that share exactly 1 edge with the current cell and we can visit a house twice or more.

Prerequisites:

  • DP and formation of DP transition relation
  • Bitmasking in DP
  • Traveling Salesman problem

Examples:

Input:

bitmasking---------and---------dynamic---------programming---------travelling---------salesman---------problem---------1

Output: 22

Input:

bitmasking---------and---------dynamic---------programming---------travelling---------salesman---------problem---------2

Output: 12

Approach - Using BFS + BitMasking

If we treat each house as a node, the problem can be viewed as finding the minimum steps required to visit all nodes. In this problem, each house is a node in a graph, and we need to find the shortest path to visit all of them. The key challenge is that the same node (house) can be visited multiple times, but only if it allows us to visit more houses. Simply revisiting a node with the same number of visited houses is redundant.

Since we are working with a grid and the cost to move to an adjacent cell is always one, we can use BFS to find the shortest path. However, since we can move in all four directions, the same node might be revisited multiple times. This means we cannot simply use a visited array as in standard BFS. Instead, we need a way to uniquely represent each state.
To address this, we assign each house a unique houseId. Once we have the houseId, we can track house visits by setting the corresponding bits in a bitmask. The bitmask represents the set of visited houses, and the coordinates (x, y) track the current position. This allows us to mark each state uniquely. Revisiting a node is only valid if it results in a different set of visited houses.

C++
// C++ program to find minimum step required to // visit all houses.  #include <bits/stdc++.h> using namespace std;  int findMinSteps(vector<vector<char>> &mat) {     int n = mat.size();     int m = mat[0].size();      // Transforming the grid: assign unique IDs to   	// houses and mark obstacles Grid to store houses and obstacles     vector<vector<int>> grid(n, vector<int>(m, 0));     int houseId = 0;     int startX = 0, startY = 0;      // Process the grid and assign unique IDs to houses,    	// and find starting position     for (int i = 0; i < n; i++) {         for (int j = 0; j < m; j++) {             if (mat[i][j] == '.') {                                // Open cell, marked as 0                 grid[i][j] = 0;             }             else if (mat[i][j] == '*') {                                // Assign unique ID to each house                 grid[i][j] = houseId++;             }             else if (mat[i][j] == 'o') {                                // Record starting row and column                 startX = i;                 startY = j;             }             else {                 // Mark obstacles as -1                 grid[i][j] = -1;             }         }     }      // If there are no houses to visit, return 0     if (houseId == 0)         return 0;      // BFS setup: Visited array and queue     vector<vector<vector<int>>> vis(1 << houseId,  	vector<vector<int>>(n, vector<int>(m, 0)));      // BFS queue to store (visitedMask, x, y)     queue<vector<int>> q;      // Mark the starting position as visited     vis[0][startX][startY] = 1;      // Push the initial state (0 mask, startX, startY)     q.push({0, startX, startY});      // Counter for the number of steps     int steps = 0;      // Directions for BFS traversal: right, left, down, up     int dx[] = {0, 0, 1, -1};     int dy[] = {1, -1, 0, 0};      // BFS loop     while (!q.empty()) {         int qSize = q.size();          for (int idx = 0; idx < qSize; idx++) {             auto curr = q.front();             q.pop();              // Mask of visited houses             int mask = curr[0];              // Current x position             int x = curr[1];              // Current y position             int y = curr[2];              // If all houses are visited, return           	// the steps taken             if (mask == (1 << houseId) - 1) {                 return steps;             }              // Explore all possible moves (right, left, down, up)             for (int k = 0; k < 4; k++)  {                                // Calculate new x  and y position.                 int newX = x + dx[k];                 int newY = y + dy[k];                  // Check boundaries and ensure it's not a blocked cell                 if (newX >= 0 && newX < n && newY >= 0                      && newY < m && grid[newX][newY] != -1) {                     if (mat[newX][newY] == '*') {                                                // If it's a house, update the visited mask                         int newMask = mask | (1 << grid[newX][newY]);                          // If the new state is already visited, continue                         if (vis[newMask][newX][newY])                             continue;                          // Mark the new state as visited                         vis[newMask][newX][newY] = 1;                                                // Push the new state to the queue                         q.push({newMask, newX, newY});                     }                     else {                                                // If it's a normal cell, continue without                       	// updating the mask                         if (vis[mask][newX][newY])                             continue;                          vis[mask][newX][newY] = 1;                         q.push({mask, newX, newY});                     }                 }             }         }                // Increment step count after processing all       	// states at the current level         steps++;     }      // If unable to visit all houses, return -1     return -1; }  int main() {      vector<vector<char>> mat = {{'.', '.', '.', '.', '.', '.', '.'},                                 {'.', 'o', '.', '.', '.', '*', '.'},                                 {'.', '.', '.', '.', '.', '.', '.'},                                 {'.', '*', '.', '.', '.', '*', '.'},                                 {'.', '.', '.', '.', '.', '.', '.'}};      int res = findMinSteps(mat);     cout << res << endl;      return 0; } 
Java
// Java program to find minimum step required // to visit all houses.  import java.util.*;    class  GfG {         static int findMinSteps(char[][] mat) {         int n = mat.length;         int m = mat[0].length;          // Transforming the grid: assign unique IDs to         // houses and mark obstacles Grid to store houses         // and obstacles         int[][] grid = new int[n][m];         int houseId = 0;         int startX = 0, startY = 0;          // Process the grid and assign unique IDs to houses,         // and find starting position         for (int i = 0; i < n; i++) {             for (int j = 0; j < m; j++) {                 if (mat[i][j] == '.') {                                        // Open cell, marked as 0                     grid[i][j] = 0;                 }                 else if (mat[i][j] == '*') {                                        // Assign unique ID to each house                     grid[i][j] = houseId++;                 }                 else if (mat[i][j] == 'o') {                                        // Record starting row and column                     startX = i;                     startY = j;                 }                 else {                                        // Mark obstacles as -1                     grid[i][j] = -1;                 }             }         }          // If there are no houses to visit, return 0         if (houseId == 0)             return 0;          // BFS setup: Visited array and queue         boolean[][][] vis = new boolean[1 << houseId][n][m];          // BFS queue to store (visitedMask, x, y)         Queue<int[]> q = new LinkedList<>();          // Mark the starting position as visited         vis[0][startX][startY] = true;          // Push the initial state (0 mask, startX, startY)         q.add(new int[] { 0, startX, startY });          // Counter for the number of steps         int steps = 0;          // Directions for BFS traversal: right, left, down,         // up         int[] dx = { 0, 0, 1, -1 };         int[] dy = { 1, -1, 0, 0 };          // BFS loop         while (!q.isEmpty()) {             int qSize = q.size();              for (int idx = 0; idx < qSize; idx++) {                 int[] curr = q.poll();                  // Mask of visited houses                 int mask = curr[0];                  // Current x position                 int x = curr[1];                  // Current y position                 int y = curr[2];                  // If all houses are visited, return the                 // steps taken                 if (mask == (1 << houseId) - 1) {                     return steps;                 }                  // Explore all possible moves (right, left,                 // down, up)                 for (int k = 0; k < 4; k++) {                                        // Calculate new x and y position.                     int newX = x + dx[k];                     int newY = y + dy[k];                      // Check boundaries and ensure it's not                     // a blocked cell                     if (newX >= 0 && newX < n && newY >= 0                         && newY < m                         && grid[newX][newY] != -1) {                         if (mat[newX][newY] == '*') {                                                        // If it's a house, update the                             // visited mask                             int newMask                                 = mask                                   | (1 << grid[newX][newY]);                              // If the new state is already                             // visited, continue                             if (vis[newMask][newX][newY])                                 continue;                              // Mark the new state as visited                             vis[newMask][newX][newY] = true;                                                        // Push the new state to the                             // queue                             q.add(new int[] { newMask, newX,                                               newY });                         }                         else {                                                        // If it's a normal cell,                             // continue without updating the                             // mask                             if (vis[mask][newX][newY])                                 continue;                              vis[mask][newX][newY] = true;                             q.add(new int[] { mask, newX,                                               newY });                         }                     }                 }             }                        // Increment step count after processing all             // states at the current level             steps++;         }          // If unable to visit all houses, return -1         return -1;     }      public static void main(String[] args) {                 char[][] mat             = { { '.', '.', '.', '.', '.', '.', '.' },                 { '.', 'o', '.', '.', '.', '*', '.' },                 { '.', '.', '.', '.', '.', '.', '.' },                 { '.', '*', '.', '.', '.', '*', '.' },                 { '.', '.', '.', '.', '.', '.', '.' } };          int res = findMinSteps(mat);         System.out.println(res);     } } 
Python
# Python program to find minimum step required  # to visit all houses.  from collections import deque   def findMinSteps(mat):     n = len(mat)     m = len(mat[0])      # Transforming the grid: assign unique IDs to     # houses and mark obstacles Grid to store houses and obstacles     grid = [[0] * m for _ in range(n)]     houseId = 0     startX = startY = 0      # Process the grid and assign unique IDs to houses,      # and find starting position     for i in range(n):         for j in range(m):             if mat[i][j] == '.':                                # Open cell, marked as 0                 grid[i][j] = 0             elif mat[i][j] == '*':                                # Assign unique ID to each house                 grid[i][j] = houseId                 houseId += 1             elif mat[i][j] == 'o':                                # Record starting row and column                 startX = i                 startY = j             else:                                # Mark obstacles as -1                 grid[i][j] = -1      # If there are no houses to visit, return 0     if houseId == 0:         return 0      # BFS setup: Visited array and queue     vis = [[[False] * m for _ in range(n)] for _ in range(1 << houseId)]      # BFS queue to store (visitedMask, x, y)     q = deque()      # Mark the starting position as visited     vis[0][startX][startY] = True      # Push the initial state (0 mask, startX, startY)     q.append([0, startX, startY])      # Counter for the number of steps     steps = 0      # Directions for BFS traversal: right, left, down, up     dx = [0, 0, 1, -1]     dy = [1, -1, 0, 0]      # BFS loop     while q:         qSize = len(q)          for _ in range(qSize):             curr = q.popleft()              # Mask of visited houses             mask = curr[0]              # Current x position             x = curr[1]              # Current y position             y = curr[2]              # If all houses are visited, return the             # steps taken             if mask == (1 << houseId) - 1:                 return steps              # Explore all possible moves (right, left, down, up)             for k in range(4):                                # Calculate new x and y position.                 newX = x + dx[k]                 newY = y + dy[k]                  # Check boundaries and ensure it's not a blocked cell                 if 0 <= newX < n and 0 <= newY < m and grid[newX][newY] != -1:                     if mat[newX][newY] == '*':                                                # If it's a house, update the visited mask                         newMask = mask | (1 << grid[newX][newY])                          # If the new state is already visited, continue                         if vis[newMask][newX][newY]:                             continue                          # Mark the new state as visited                         vis[newMask][newX][newY] = True                                                  # Push the new state to the queue                         q.append([newMask, newX, newY])                     else:                                                # If it's a normal cell, continue without                         # updating the mask                         if vis[mask][newX][newY]:                             continue                          vis[mask][newX][newY] = True                         q.append([mask, newX, newY])          # Increment step count after processing all states         # at the current level         steps += 1      # If unable to visit all houses,     # return -1     return -1     mat = [     ['.', '.', '.', '.', '.', '.', '.'],     ['.', 'o', '.', '.', '.', '*', '.'],     ['.', '.', '.', '.', '.', '.', '.'],     ['.', '*', '.', '.', '.', '*', '.'],     ['.', '.', '.', '.', '.', '.', '.'] ]  res = findMinSteps(mat) print(res) 
C#
// C# program to find minimum step required // to visit all houses.  using System; using System.Collections.Generic;  class GfG {           static int findMinSteps(char[, ] mat) {         int n = mat.GetLength(0);         int m = mat.GetLength(1);          // Transforming the grid: assign unique IDs to         // houses and mark obstacles Grid to store houses         // and obstacles         int[, ] grid = new int[n, m];         int houseId = 0;         int startX = 0, startY = 0;          // Process the grid and assign unique IDs to houses,         // and find starting position         for (int i = 0; i < n; i++) {             for (int j = 0; j < m; j++) {                 if (mat[i, j] == '.') {                                        // Open cell, marked as 0                     grid[i, j] = 0;                 }                 else if (mat[i, j] == '*') {                                        // Assign unique ID to each house                     grid[i, j] = houseId;                     houseId++;                 }                 else if (mat[i, j] == 'o') {                                        // Record starting row and column                     startX = i;                     startY = j;                 }                 else {                                        // Mark obstacles as -1                     grid[i, j] = -1;                 }             }         }          // If there are no houses to visit, return 0         if (houseId == 0)             return 0;          // BFS setup: Visited array and queue         bool[, , ] vis = new bool[1 << houseId, n, m];          // BFS queue to store (visitedMask, x, y)         Queue<int[]> q = new Queue<int[]>();          // Mark the starting position as visited         vis[0, startX, startY] = true;          // Push the initial state (0 mask, startX, startY)         q.Enqueue(new int[] { 0, startX, startY });          // Counter for the number of steps         int steps = 0;          // Directions for BFS traversal: right, left, down,         // up         int[] dx = { 0, 0, 1, -1 };         int[] dy = { 1, -1, 0, 0 };          // BFS loop         while (q.Count > 0) {             int qSize = q.Count;              for (int idx = 0; idx < qSize; idx++) {                 int[] curr = q.Dequeue();                  // Mask of visited houses                 int mask = curr[0];                  // Current x position                 int x = curr[1];                  // Current y position                 int y = curr[2];                  // If all houses are visited, return the                 // steps taken                 if (mask == (1 << houseId) - 1) {                     return steps;                 }                  // Explore all possible moves (right, left,                 // down, up)                 for (int k = 0; k < 4; k++) {                                        // Calculate new x and y position.                     int newX = x + dx[k];                     int newY = y + dy[k];                      // Check boundaries and ensure it's not                     // a blocked cell                     if (newX >= 0 && newX < n && newY >= 0                         && newY < m                         && grid[newX, newY] != -1) {                         if (mat[newX, newY] == '*') {                                                        // If it's a house, update the                             // visited mask                             int newMask                                 = mask                                   | (1 << grid[newX, newY]);                              // If the new state is already                             // visited, continue                             if (vis[newMask, newX, newY])                                 continue;                              // Mark the new state as visited                             vis[newMask, newX, newY] = true;                                                        // Push the new state to the                             // queue                             q.Enqueue(new int[] {                                 newMask, newX, newY });                         }                         else {                                                        // If it's a normal cell,                             // continue without updating the                             // mask                             if (vis[mask, newX, newY])                                 continue;                              vis[mask, newX, newY] = true;                             q.Enqueue(new int[] {                                 mask, newX, newY });                         }                     }                 }             }              // Increment step count after processing all             // states at the current level             steps++;         }          // If unable to visit all houses, return -1         return -1;     }      static void Main(string[] args) {         char[, ] mat             = { { '.', '.', '.', '.', '.', '.', '.' },                 { '.', 'o', '.', '.', '.', '*', '.' },                 { '.', '.', '.', '.', '.', '.', '.' },                 { '.', '*', '.', '.', '.', '*', '.' },                 { '.', '.', '.', '.', '.', '.', '.' } };          int res = findMinSteps(mat);         Console.WriteLine(res);     } } 
JavaScript
 // JavaScript program to find minimum step required  // to visit all houses.   function findMinSteps(mat) {     const n = mat.length;     const m = mat[0].length;      // Transforming the grid: assign unique IDs to houses     // and mark obstacles Grid to store houses and obstacles     const grid         = Array.from({length : n}, () => Array(m).fill(0));     let houseId = 0;     let startX = 0, startY = 0;      // Process the grid and assign unique IDs to houses, and     // find starting position     for (let i = 0; i < n; i++) {         for (let j = 0; j < m; j++) {             if (mat[i][j] === ".") {                              // Open cell, marked as 0                 grid[i][j] = 0;             }             else if (mat[i][j] === "*") {                              // Assign unique ID to each house                 grid[i][j] = houseId;                 houseId++;             }             else if (mat[i][j] === "o") {                              // Record starting row and column                 startX = i;                 startY = j;             }             else {                              // Mark obstacles as -1                 grid[i][j] = -1;             }         }     }      // If there are no houses to visit, return 0     if (houseId === 0)         return 0;      // BFS setup: Visited array and queue     const vis = Array.from(         {length : 1 << houseId},         () => Array.from({length : n},                          () => Array(m).fill(false)));      // BFS queue to store (visitedMask, x, y)     const queue = [];      // Mark the starting position as visited     vis[0][startX][startY] = true;      // Push the initial state (0 mask, startX, startY)     queue.push([ 0, startX, startY ]);      // Counter for the number of steps     let steps = 0;      // Directions for BFS traversal: right, left, down, up     const dx = [ 0, 0, 1, -1 ];     const dy = [ 1, -1, 0, 0 ];      // BFS loop     while (queue.length > 0) {         const qSize = queue.length;          for (let idx = 0; idx < qSize; idx++) {             const [mask, x, y] = queue.shift();              // If all houses are visited, return the steps             // taken             if (mask === (1 << houseId) - 1) {                 return steps;             }              // Explore all possible moves (right, left,             // down, up)             for (let k = 0; k < 4; k++) {                              // Calculate new x and y position.                 const newX = x + dx[k];                 const newY = y + dy[k];                  // Check boundaries and ensure it's not a                 // blocked cell                 if (newX >= 0 && newX < n && newY >= 0                     && newY < m                     && grid[newX][newY] !== -1) {                     if (mat[newX][newY] === "*") {                                              // If it's a house, update the                         // visited mask                         const newMask                             = mask                               | (1 << grid[newX][newY]);                          // If the new state is already                         // visited, continue                         if (vis[newMask][newX][newY])                             continue;                          // Mark the new state as visited                         vis[newMask][newX][newY] = true;                                                  // Push the new state to the queue                         queue.push([ newMask, newX, newY ]);                     }                     else {                                              // If it's a normal cell, continue                         // without updating the mask                         if (vis[mask][newX][newY])                             continue;                          vis[mask][newX][newY] = true;                         queue.push([ mask, newX, newY ]);                     }                 }             }         }          // Increment step count after processing all states         // at the current level         steps++;     }      // If unable to visit all houses, return -1     return -1; }  const mat = [     [ ".", ".", ".", ".", ".", ".", "." ],     [ ".", "o", ".", ".", ".", "*", "." ],     [ ".", ".", ".", ".", ".", ".", "." ],     [ ".", "*", ".", ".", ".", "*", "." ],     [ ".", ".", ".", ".", ".", ".", "." ] ];  const res = findMinSteps(mat); console.log(res); 

Output
8 

Time Complexity: O(2^totalHouse*n*m) where n and m are number of rows and columns of matrix and totalHouse is the number of house in matrix.
Auxilary Space: O(2^totalHouse*n*m)


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Bitmasking and Dynamic Programming | Travelling Salesman Problem

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Article Tags :
  • Bit Magic
  • Dynamic Programming
  • DSA
Practice Tags :
  • Bit Magic
  • Dynamic Programming

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    Given a positive integer n, the task is to find if it is a power of 2 or not.Examples: Input : n = 16Output : YesExplanation: 24 = 16Input : n = 42Output : NoExplanation: 42 is not a power of 2Input : n = 1Output : YesExplanation: 20 = 1Approach 1: Using Log - O(1) time and O(1) spaceThe idea is to
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    Find position of the only set bit
    Given a number n containing only 1 set bit in its binary representation, the task is to find the position of the only set bit. If there are 0 or more than 1 set bits, then return -1. Note: Position of set bit '1' should be counted starting with 1 from the LSB side in the binary representation of the
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    Check for Integer Overflow
    Given two integers a and b. The task is to design a function that adds two integers and detects overflow during the addition. If the sum does not cause an overflow, return their sum. Otherwise, return -1 to indicate an overflow.Note: You cannot use type casting to a larger data type to check for ove
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    Find XOR of two number without using XOR operator
    Given two integers, the task is to find XOR of them without using the XOR operator.Examples : Input: x = 1, y = 2Output: 3Input: x = 3, y = 5Output: 6Approach - Checking each bit - O(log n) time and O(1) spaceA Simple Solution is to traverse all bits one by one. For every pair of bits, check if both
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    Check if two numbers are equal without using arithmetic and comparison operators
    Given two numbers, the task is to check if two numbers are equal without using Arithmetic and Comparison Operators or String functions. Method 1 : The idea is to use XOR operator. XOR of two numbers is 0 if the numbers are the same, otherwise non-zero. C++ // C++ program to check if two numbers // a
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    Detect if two integers have opposite signs
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    Swap Two Numbers Without Using Third Variable
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