Problem Statement
In this problem, you are given two integers m and n, representing the dimensions (height and width, respectively) of a 0-indexed 2D array board. Additionally, a pair of integers (r, c) specifies the starting position of a knight on this board. The core challenge is to determine a sequence of moves for the knight such that it visits every cell on the board exactly once, starting from the initial position (r, c). The cell values of the resultant array board should reflect the order in which the cells are visited, starting with 0 at the initial position of the knight.
A knight's move in chess allows it to move in an L-shape: two cells in one direction and one cell in the perpendicular direction. Formally, a knight at position (r1, c1) can move to (r2, c2) if the movement satisfies the conditions 0 <= r2 <= m - 1, 0 <= c2 <= n - 1, and the minimum absolute difference in one dimension is 1, while the maximum in the other dimension is 2, or vice versa.
The objective is to output the 2D array board with the cells populated in the sequence of the knight's visits.
Examples
Example 1
Input:
m = 1, n = 1, r = 0, c = 0
Output:
[[0]]
Explanation:
There is only one cell. The knight starts there and the tour is complete.
Example 2
Input:
m = 3, n = 4, r = 0, c = 0
Output:
[[0, 5, 10, 15], [9, 14, 1, 6], [4, 11, 8, 13]]
Explanation:
This is one valid knight's tour from position (0, 0). Each cell shows the step number in the tour.
Constraints
1 <= m, n <= 50 <= r < m0 <= c < n- It is guaranteed that at least one valid knight's tour exists from the starting position.
Approach and Intuition
Given the unique movements of a knight in chess and the small potential sizes of the board (as defined by maximum dimensions 5x5), we can visually or programmatically determine a path that visits all cells once. Here's how one might think about tackling this:
- Begin with the knight's initial position as the first entry in the matrix.
- Knowing the movement rules of a knight, generate possible moves from any position
(r, c). These could include positions like(r + 2, c + 1),(r + 2, c - 1),(r - 2, c + 1),(r - 2, c - 1),(r + 1, c + 2),(r + 1, c - 2),(r - 1, c + 2), and(r - 1, c - 2). Each of these moves must also validate whether they fall within the boundaries of the board and have not been previously visited. - Utilize a search technique, such as Depth-First Search (DFS), possibly with heuristics or backtracking, to explore all paths that proceed from the starting knight position. The goal is to identify a sequence that successfully visits each cell exactly once.
- Continuously check if the number of moves made matches the number of cells, indicating all cells have been covered without repetitions.
- The complexity and feasibility of this chess puzzle-like problem are reasonable due to the size restrictions (
1 <= m, n <= 5), allowing for backtracking solutions without significant performance concerns.
The steps and the possible paths would be recursively determined, and the algorithm ensures we always find at least one valid sequence, as guaranteed by the constraints. Thus, such an approach optimally leverages the knight's unique movements and the limited size of the board to find a solution.
Solutions
- C++
- Java
- Python
class KnightsTourSolver {
public:
vector<vector<int>> generateTour(int x_dim, int y_dim, int start_x, int start_y) {
vector<vector<int>> board(x_dim, vector<int>(y_dim, 0));
board[start_x][start_y] = -1;
recursiveSolve(x_dim, y_dim, start_x, start_y, board, 1);
board[start_x][start_y] = 0;
return board;
}
private:
int moves[8][2] = {{-1, -2}, {-2, -1}, {-1, 2}, {-2, 1},
{1, -2}, {2, -1}, {1, 2}, {2, 1}};
bool recursiveSolve(int width, int height, int row, int col,
vector<vector<int>>& board, int step) {
if (step == width * height) return true;
auto feasibleMoves = getMovesAccordingToWarnsdorff(board, row, col);
for (auto& move : feasibleMoves) {
int newRow = row + moves[move.second][0];
int newCol = col + moves[move.second][1];
if (!isValidPosition(board, newRow, newCol)) continue;
board[newRow][newCol] = step;
if (recursiveSolve(width, height, newRow, newCol, board, step + 1)) {
return true;
}
board[newRow][newCol] = 0;
}
return false;
}
vector<pair<int, int>> getMovesAccordingToWarnsdorff(
vector<vector<int>>& board, int row, int col) {
vector<pair<int, int>> moveOptions;
for (int i = 0; i < 8; i++) {
int newRow = row + moves[i][0];
int newCol = col + moves[i][1];
int score = countPossibleMoves(board, newRow, newCol);
moveOptions.push_back({score, i});
}
sort(moveOptions.begin(), moveOptions.end());
return moveOptions;
}
int countPossibleMoves(vector<vector<int>>& board, int row, int col) {
int movesCount = 0;
for (int i = 0; i < 8; i++) {
int newRow = row + moves[i][0];
int newCol = col + moves[i][1];
if (isValidPosition(board, newRow, newCol)) movesCount++;
}
return movesCount;
}
bool isValidPosition(vector<vector<int>>& board, int row, int col) {
return row >= 0 && col >= 0 && row < board.size() &&
col < board[0].size() && board[row][col] == 0;
}
};
The Knight’s Tour problem is a classic example of using backtracking to solve a complex puzzle in C++. In the provided solution, a KnightsTourSolver class is designed to find a path for a knight to visit every cell on an x_dim by y_dim chessboard exactly once, starting from a given position (start_x, start_y).
The main function,
generateTour, initializes a two-dimensional board with zero values and sets the starting position of the knight with a special marker (e.g., -1). The function then attempts to complete the tour using a recursive approachrecursiveSolve, after which it resets the start position and returns the board representing the tour.The recursive function
recursiveSolveexplores all possible knight moves. It adheres to the chess rule for knight moves (shaped like an "L") and ensures moves do not go outside the board or revisit an already visited cell. Notably, it incorporates Warnsdorff's rule by prioritizing moves which have the least onward moves available, making the backtracking process more efficient.Helper function
getMovesAccordingToWarnsdorffcalculates these possibilities. It generates score-based move options, sorting them to promote moves with fewer continuations, decreasing the likeliness of getting stuck.Another helper,
isValidPosition, ensures the moves stay within the bounds of the board and targets only unvisited cells, crucial for maintaining the validity of each step in the tour.
Using these well-structured methods, the program meticulously builds a solution path for the knight, adhering strictly to the rules of its movement and optimizing the process with strategic move ordering. Upon obtaining a valid tour, the solver provides a board filled with step numbers representing the knight's path, or an unaltered board if no tour is possible. This makes it possible to visually trace the knight's route across the chessboard.
class KnightTourSolver {
int[][] knightDirections = {
{ -1, -2 }, { -2, -1 },
{ -1, 2 }, { -2, 1 },
{ 1, -2 }, { 2, -1 },
{ 1, 2 }, { 2, 1 },
};
public int[][] calculateKnightTour(int boardHeight, int boardWidth, int startRow, int startCol) {
int[][] board = new int[boardHeight][boardWidth];
board[startRow][startCol] = -1;
solveTour(boardHeight, boardWidth, startRow, startCol, board, 1);
board[startRow][startCol] = 0;
return board;
}
private boolean solveTour(
int height,
int width,
int row,
int col,
int[][] board,
int moveNum
) {
if (moveNum == height * width) return true;
List<int[]> moves = prioritizeMoves(board, row, col);
for (int[] move : moves) {
int newRow = row + knightDirections[move[1]][0];
int newCol = col + knightDirections[move[1]][1];
if (!isLegalMove(board, newRow, newCol)) continue;
board[newRow][newCol] = moveNum;
if (solveTour(height, width, newRow, newCol, board, moveNum + 1)) {
return true;
}
board[newRow][newCol] = 0;
}
return false;
}
private List<int[]> prioritizeMoves(int[][] board, int row, int col) {
List<int[]> moves = new ArrayList<>();
for (int i = 0; i < knightDirections.length; i++) {
int nextRow = row + knightDirections[i][0];
int nextCol = col + knightDirections[i][1];
int futureMovesCount = countValidMoves(board, nextRow, nextCol);
moves.add(new int[] { futureMovesCount, i });
}
Collections.sort(moves, (a, b) -> a[0] - b[0]);
return moves;
}
private int countValidMoves(int[][] board, int row, int col) {
int count = 0;
for (int[] direction : knightDirections) {
int nextRow = row + direction[0];
int nextCol = col + direction[1];
if (isLegalMove(board, nextRow, nextCol)) count++;
}
return count;
}
private boolean isLegalMove(int[][] board, int row, int col) {
return (
row >= 0 &&
col >= 0 &&
row < board.length &&
col < board[0].length &&
board[row][col] == 0
);
}
}
The Knight’s Tour problem involves finding a sequence of moves for a knight on a chessboard such that the knight visits every square exactly once. Using Java, the provided solution utilizes a backtracking approach complemented by a heuristic to efficiently solve the problem.
- Encapsulated within a
KnightTourSolverclass, this Java implementation defines a methodcalculateKnightTourwhich initiates the tour from a specified starting position on the board. - A crucial method,
solveTour, recursively attempts to place the knight in a legitimate position, considering all possible moves from the current position. It checks if the current path leads to a complete solution before backtracking. - The heuristic employed is the Warnsdorff’s rule, which prioritizes moves based on the count of onward moves from the destination cell. This heuristic significantly reduces the solution space by choosing paths that are likely to lead to success earlier.
prioritizeMovescalculates possible moves from the current cell and uses Java’s sorting mechanisms to rank these based on the number of subsequent legal moves, thus implementing Warnsdorff’s rule.- The method
isLegalMoveensures the move is within the bounds of the board and to an unvisited square.
Each part of the code integrates to create a solution that efficiently navigates the complexities of the Knight’s Tour, providing a structured approach to exploring potential paths on the board. The use of a backtracking paired with a heuristic offers a practical balance between exhaustive search and intelligent pruning, making this approach suitable for larger boards.
class Solution:
def knightJourney(self, width, height, start_row, start_col):
# Knight movement offsets
moves = [
(2, 1), (1, 2), (-1, 2), (-2, 1),
(-2, -1), (-1, -2), (1, -2), (2, -1)
]
board = [[0] * height for _ in range(width)]
board[start_row][start_col] = -1
def _explore(current_r, current_c, step_number):
if step_number == width * height:
return True
possible_moves = _sort_moves_by_options(current_r, current_c)
for score, idx in possible_moves:
next_r, next_c = current_r + moves[idx][0], current_c + moves[idx][1]
if not _is_legal(next_r, next_c):
continue
board[next_r][next_c] = step_number
if _explore(next_r, next_c, step_number + 1):
return True
board[next_r][next_c] = 0
return False
def _sort_moves_by_options(r, c):
move_options = []
for i in range(8):
nr, nc = r + moves[i][0], c + moves[i][1]
if _is_legal(nr, nc):
score = sum(
_is_legal(nr + d[0], nc + d[1]) for d in moves
)
move_options.append((score, i))
return sorted(move_options)
def _is_legal(r, c):
return 0 <= r < width and 0 <= c < height and board[r][c] == 0
_explore(start_row, start_col, 1)
board[start_row][start_col] = 0
return board
This Python solution tackles the Knight’s Tour problem, where the task is to determine a sequence of moves for a knight on a chessboard such that the knight visits every square exactly once. The function knightJourney(width, height, start_row, start_col) initializes the board and employs a backtracking algorithm to explore potential tours.
- Initialize an empty chessboard with dimensions specified by
widthandheight, marking the starting position with-1. - Define the legal movements of a knight on the chessboard using a list of tuples
moves. - Implement a helper function
_explore()to recursively track the knight's path:- Terminate successfully if the knight completes the tour by visiting all squares.
- Heuristically prioritize the knight's next moves using
_sort_moves_by_options()to favor movements leading to less accessible squares. This employs Warnsdorff's rule for efficiency. - Continue exploring from the next valid position. If a dead end is reached, backtrack by resetting the square and exploring alternate paths.
_sort_moves_by_options(r, c)computes possible moves from a given position, ranking them based on the number of onward moves available from the subsequent positions. This heuristic assists in reducing exploration of branch paths likely to lead to dead ends early on in the search._is_legal(r, c)checks if a position on the board is legal for the knight to move to, ensuring it's within bounds and not previously visited.
The algorithm prioritizes move efficiency and uses backtracking to handle dead-ends, resetting paths as needed. When a valid Knight's Tour is found, the function returns the board's final state showing the path followed by the knight. If no tour is possible, the returning board configuration will reflect that fact.
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