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The Knight's Tour Problem (using Backtracking Algorithm)

Jun 30, 2023
6 Minute Read
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By Vedanti Kshirsagar

Ever wondered how a computer playing as a chess player improves its algorithm to defeat you in the game? In this article, we will learn about the Knight's Tour Problem, the various ways to solve it along with their time and space complexities. So, let's get started!

What is Knight's Tour Problem?

Just for a reminder, the knight is a piece in chess that usually looks like a horse and moves in an L-shaped pattern. This means it will first move two squares in one direction and then one square in a perpendicular direction.

The Knight's Tour problem is about finding a sequence of moves for the knight on a chessboard such that it visits every square on the board exactly one time. It is a type of Hamiltonian path problem in graph theory, where the squares represent the vertices and the knight's moves represent the edges of the graph.

This problem has fascinated many mathematicians for centuries, but the solutions they found were very difficult. The simple solution will be to find every conceivable configuration and selecting the best one is one technique to solve this problem. But that will take a load of time.

One popular solution to solve the Knight's tour problem is Warnsdorff's rule, which involves choosing the next move that leads to a square with the fewest available moves. There is also a backtracking approach.

But first moving to all that, let's take a quick understanding of the Hamiltonian path problem.

Hamiltonian Path Problem

The Hamiltonian path problem is a well-known problem in graph theory that asks whether a given graph contains a path that visits every vertex exactly once.

A path that visits every vertex exactly once is called a Hamiltonian path, and a graph that contains a Hamiltonian path is called a Hamiltonian graph.

Let's take an example of the Hamiltonian path problem. Suppose we have a graph with five vertices and the following edges:

This graph can be represented as:

Knight's Tour Backtracking Algorithm

The backtracking algorithm works by exploring all possible moves for the knight, starting from a given square, and backtracking to try different moves if it reaches a dead end.

Here's the basic outline of the backtracking algorithm to solve the Knight's tour problem:

Choose a starting square for the knight on the chessboard.
Mark the starting square as visited.
For each valid move from the current square, make the move and recursively repeat the process for the new square.
If all squares on the chessboard have been visited, we have found a solution. Otherwise, undo the last move and try a different move.
If all moves have been tried from the current square and we have not found a solution, backtrack to the previous square and try a different move from there.
If we have backtracked to the starting square and tried all possible moves without finding a solution, there is no solution to the problem.

We have given the full C++ program for Backtracking Algorithm to solve Knight's Tour Problem below:

Check the image below before we explain the code:

In this implementation, we first define a function validMoves which takes the current row and column of the knight as input and returns a vector of pairs representing all the valid moves that the knight can make from that position.

The solve function is a recursive function that takes the current row and column of the knight, as well as the current move number, as input. We mark the current square as visited by setting board[row][column] it to the current move number, and then we recursively try all possible moves from the current position.

If we reach the last move, then we found a solution and return true. If no valid move is found from the current position, we backtrack by setting the current square to 0 and returning false.

In the main function, we start the solve function at a specified starting position and then output the number of moves it took to find a solution and print the final chessboard with the solution.

Time & Space Complexity for Backtracking

The backtracking algorithm used to solve the Knight's Tour problem has an exponential time complexity. The number of possible paths for the knight grows very quickly as the size of the chessboard increases, which means that the time taken to explore all possible paths grows exponentially.

The exact time complexity of the Knight's Tour Backtracking algorithm is O(8^(n^2)), where n is the size of the chessboard. This is because each move has a maximum of 8 possible directions, and we have to explore all possible moves until we find a solution.

The space complexity of the backtracking algorithm is O(n^2), where n is the size of the chessboard. So, we can say that the backtracking algorithm is efficient for smaller chessboards.

Warnsdorff's Rule

Warndorff's rule is a heuristic greedy algorithm used to solve this problem. It tries to move the knight to the square with the fewest valid moves, hoping that this will lead to a solution.

Here's an overview of how Warndorff's rule algorithm works:

Start with a random square on the chessboard.
From the current square, consider all possible moves and count the number of valid moves from each adjacent square.
Move to the square with the lowest number of valid moves. In case of a tie, move to the square with the lowest number of valid moves from its adjacent squares.
Repeat steps 2 and 3 until all squares on the chessboard have been visited.

Here is the pseudocode for Warndorff's rule algorithm:

The time complexity of Warndorff's rule algorithm is O(n^2 log n), where n is the size of the chessboard. This is because we have to visit each square once, and for each square, we have to compute the number of valid moves from each adjacent square. The space complexity of the algorithm is O(n^2) since we need to store the chessboard and the current position of the knight.

Overall, The Knight's Tour problem is related to chess, and solving it can help chess players improve their strategy and become better at the game. In the real world, you can also use it to design efficient algorithms for finding the shortest path between two points in a network.

Now we know The Knight's Tour Problem and its solutions using Backtracking and Warnsdorff's Rule. It has several applications in various fields such as Game theory, Network Routing etc, making it an important problem to study in computer science and mathematics.

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Knight’s Tour Problem

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The Knight's Tour Problem is one of the famous problem in which we have the knight on a chessboard. The knight is supposed to visit every square exactly once.

Following it we have two cases:

Closed Tour : The knight ends on a square that is one move away from the beginning square. This means that if it continues to move, with the same path that it has followed till now, it will be able to traverse all squares again.
Open Tour : The knight ends on any other square. This problem is solved on a n x n chessboard. "n" could have any value. This problem is derived from the Hamiltonian Path Problem in Graph Theory .

Hamiltonian Path Problem is focused on finding out if there is a path or route in undirected graph from the beginning to ending node. However, it can gain a lot of complexity even on little increase in its size. Hamiltonian Path is the one that visits each vertex of the graph only once. A grap h with Hamiltonian Path in it is called traceable graph.

However, there are ways to solve this problem in linear time. Consider the below image, let the diamond represent the "knight" and the tick marks represent its possible moves at given position. A "knight" can move two squares in cardinal direction then one square in orthogonal direction.

There are many variants possible for this problem like having two knights and so on. In this article we will explore the basic problem with one knight and how it can cover all squares without revisiting them.

One way to solve this problem is Naive Approach, to find all possible configurations and choose the correct one. Since, this would take much longer if the value of n in n x n board is large.

Backtracking

Other technique that can be used for this problem is Backtracking . It is optimized version of our naive technique. Backtracking is the technique in which we try all possible solutions or combinations till we find the one that is correct. In this each combination is tried only once.

It incrementally makes the solution using recursion. We consider on step at a time. If they do not satisfy the constraints then we remove those particular steps.

Time Complexity: O(n * n) for traversing the n x n squares. Implementation:

Explanation:

The function called in main() would be ans().
It declares a sol array representing the chess board of n x n size.
Then we initialise it with -1 values. Then we define the next values of x and y coordinates in array x and array y.
Now, we call function aux and explore all tours and check if the solution exists.
If it exists then we print the solution. The aux function:
This is the recursive function to find the final solution.
In this we try all possible moves from x and y. We also call function range to check if the squares are in n x n configuration.
We essentially use backtracking to solve this problem.

We can have much better approaches as well to solve this problem:

Warnsdorff’s algorithm:

In this approach we can start from any initial square on the chess board. Then, we will move to an adjacent, unvisited square with minimum unvisited adjacent squares.

Set a square as the initial position to start from.
Mark it with current move number starting with "1".
Make a set of all positions that can be visited from initial position.
Mark the square out of the set of possible positions that can be visited with minimum accessibility from initial position as next move number.
Following the above steps, return the marked board with move numbers with which it has been visited.

In the above code, in find function, we initialized an array of size n x n with values -1.

Then, we initialized x1 and y1 with random values.

Then we copied those values in variables x and y. Now, current positions are same as initial positions.

Then, we mark our first move at location y * n + x.

In the next for loop, we pick next points for our Warnsdorff’s algorithm. We call next function:

We will try all adjacent squares starting from a random adjacent value with minimum degree.
We initialize start with a random value
If now next cell is found then we return false.
Then, we store coordinates in x and y of next point in variables nx and ny.
Then, we mark our next move on the array at location ny * n + nx and update next points.

Now, we come back into function find and then neighbor function is called.

We will check the neighboring squares. If the last square is one move away from the initial location of square then its a closed tour else open tour.

The degree function returns the number of empty squares adjacent to x,y

The range function keeps us within the board and isempty function checks if the square is valid and empty or not.

Now, we again come back to out find function and call print function to print all moves of the knight.

This way,we can implement this algorithm and find our best possible solution.

Neural Networks:

We can even use neural networks to solve this problem. It will be much helpful in the cases where n has much larger value. Each and every possible knight's move is represented by a neuron. Each neuron can have two values : "active" or "inactive". If it is "active" then it is part of our solution else not. This way, we can use machine learning to find solutions to much more complex variants of this problem including large values for "n" and multiple knights.

Using all the above approaches we can find out the best possible solution for our Knight's Tour Problem . Happy Coding!

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The Knight’s tour problem

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Backtracking is an algorithmic paradigm that tries different solutions until finds a solution that “works”. Problems which are typically solved using backtracking technique have the following property in common. These problems can only be solved by trying every possible configuration and each configuration is tried only once. A Naive solution for these problems is to try all configurations and output a configuration that follows given problem constraints. Backtracking works in an incremental way and is an optimization over the Naive solution where all possible configurations are generated and tried.

For example, consider the following Knight’s Tour problem. The knight is placed on the first block of an empty board and, moving according to the rules of chess, must visit each square exactly once.

Path followed by Knight to cover all the cells

Naive Algorithm for Knight’s tour The Naive Algorithm is to generate all tours one by one and check if the generated tour satisfies the constraints.

Backtracking works in an incremental way to attack problems. Typically, we start from an empty solution vector and one by one add items (Meaning of item varies from problem to problem. In context of Knight’s tour problem, an item is a Knight’s move). When we add an item, we check if adding the current item violates the problem constraint, if it does then we remove the item and try other alternatives. If none of the alternatives work out then we go to the previous stage and remove the item added in the previous stage. If we reach the initial stage back then we say that no solution exists. If adding an item doesn’t violate constraints then we recursively add items one by one. If the solution vector becomes complete then we print the solution.

Backtracking Algorithm for Knight’s tour Following is the Backtracking algorithm for Knight’s tour problem.

Following are implementations for Knight’s tour problem. It prints one of the possible solutions in 2D matrix form. Basically, the output is a 2D 8*8 matrix with numbers from 0 to 63 and these numbers show steps made by Knight.

Note that Backtracking is not the best solution for the Knight’s tour problem. See below article for other better solutions. The purpose of this post is to explain Backtracking with an example. Warnsdorff’s algorithm for Knight’s tour problem

References: http://see.stanford.edu/materials/icspacs106b/H19-RecBacktrackExamples.pdf http://www.cis.upenn.edu/~matuszek/cit594-2009/Lectures/35-backtracking.ppt http://mathworld.wolfram.com/KnightsTour.html http://en.wikipedia.org/wiki/Knight%27s_tour

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A Knight’s Tour

The “ knight’s tour ” is a classic problem in graph theory, first posed over 1,000 years ago and pondered by legendary mathematicians including Leonhard Euler before finally being solved in 1823. We will use the knight’s tour problem to illustrate a second common graph algorithm called depth first search.

The knight’s tour puzzle is played on a chess board with a single chess piece, the knight. The object of the puzzle is to find a sequence of moves that allow the knight to visit every square on the board exactly once, like so:

One such sequence is called a “tour.” The upper bound on the number of possible legal tours for an eight-by-eight chessboard is known to be 1 . 3 0 5 × 1 0 3 5 1.305 \times 10^{35} 1 . 3 0 5 × 1 0 3 5 ; however, there are even more possible dead ends. Clearly this is a problem that requires some real brains, some real computing power, or both.

Once again we will solve the problem using two main steps:

Represent the legal moves of a knight on a chessboard as a graph.
Use a graph algorithm to find a path where every vertex on the graph is visited exactly once.

Building the Knight’s Tour Graph

To represent the knight’s tour problem as a graph we will use the following two ideas: Each square on the chessboard can be represented as a node in the graph. Each legal move by the knight can be represented as an edge in the graph.

We will use a Python dictionary to hold our graph, with the keys being tuples of coordinates representing the squares of the board, and the values being sets representing the valid squares to which a knight can move from that square.

To build the full graph for an n-by-n board we can use the Python function shown below. The build_graph function makes one pass over the entire board. At each square on the board the build_graph function calls a helper generator, legal_moves_from , to generate a list of legal moves for that position on the board. All legal moves are then made into undirected edges of the graph by adding the vertices appropriately to one anothers sets of legal moves.

The legal_moves_from generator below takes the position of the knight on the board and yields any of the eight possible moves that are still on the board.

The illustration below shows the complete graph of possible moves on an eight-by-eight board. There are exactly 336 edges in the graph. Notice that the vertices corresponding to the edges of the board have fewer connections (legal moves) than the vertices in the middle of the board. Once again we can see how sparse the graph is. If the graph was fully connected there would be 4,096 edges. Since there are only 336 edges, the adjacency matrix would be only 8.2 percent full.

All legal moves for a knight on an 8x8 chessboard

Implementing Knight’s Tour

The depth first exploration of the graph is exactly what we need in order to find a path that has exactly 63 edges. We will see that when the depth first search algorithm finds a dead end (a place in the graph where there are no more moves possible) it backs up the tree to the next deepest vertex that allows it to make a legal move.

The find_solution_for function takes just two arguments: a board_size argument and a heuristic function, which you should ignore for now but to which we will return.

It then constructs a graph using the build_graph function described above, and for each vertex in the graph attempts to traverse depth first by way of the traverse function.

The traverse function is a little more interesting. It accepts a path, as a list of coordinates, as well as the vertex currently being considered. If the traversal has proceeded deep enough that we know that every square has been visited once, then we return the full path traversed.

Otherwise, we use our graph to look up the legal moves from the current vertex, and exclude the vertices that we know have already been visited, to determine the vertices that are yet_to_visit . At this point we recursively call traverse with each of the vertices to visit, along with the path to reach that vertex including the current vertex. If any of the recursive calls return a path, then that path is the return value of the outer call, otherwise we return None.

Let’s look at a simple example of an equivalent of this traverse function in action.

It is remarkable that our choice of data structure and algorithm has allowed us to straightforwardly solve a problem that remained impervious to thoughtful mathematical investigation for centuries.

With some modification, the algorithm can also be used to discover one of a number of “closed” (circular) tours, which can therefore be started at any square of the board:

Knight’s Tour Analysis

There is one last interesting topic regarding the knight’s tour problem, then we will move on to the general version of the depth first search. The topic is performance. In particular, our algorithm is very sensitive to the method you use to select the next vertex to visit. For example, on a five-by-five board you can produce a path in about 1.5 seconds on a reasonably fast computer. But what happens if you try an eight-by-eight board? In this case, depending on the speed of your computer, you may have to wait up to a half hour to get the results! The reason for this is that the knight’s tour problem as we have implemented it so far is an exponential algorithm of size O ( k N ) O(k^N) O ( k N ) , where N is the number of squares on the chess board, and k is a small constant. The diagram below can help us visualize why this is so.

The root of the tree represents the starting point of the search. From there the algorithm generates and checks each of the possible moves the knight can make. As we have noted before the number of moves possible depends on the position of the knight on the board. In the corners there are only two legal moves, on the squares adjacent to the corners there are three and in the middle of the board there are eight. The diagram below shows the number of moves possible for each position on a board. At the next level of the tree there are once again between 2 and 8 possible next moves from the position we are currently exploring. The number of possible positions to examine corresponds to the number of nodes in the search tree.

Number of possible moves for each square

We have already seen that the number of nodes in a binary tree of height N is 2 N + 1 − 1 2^{N+1}-1 2 N + 1 − 1 . For a tree with nodes that may have up to eight children instead of two the number of nodes is much larger. Because the branching factor of each node is variable, we could estimate the number of nodes using an average branching factor. The important thing to note is that this algorithm is exponential: k N + 1 − 1 k^{N+1}-1 k N + 1 − 1 , where k k k is the average branching factor for the board. Let’s look at how rapidly this grows! For a board that is 5x5 the tree will be 25 levels deep, or N = 24 counting the first level as level 0. The average branching factor is k = 3 . 8 k = 3.8 k = 3 . 8 So the number of nodes in the search tree is 3 . 8 2 5 − 1 3.8^{25}-1 3 . 8 2 5 − 1 or 3 . 1 2 × 1 0 1 4 3.12 \times 10^{14} 3 . 1 2 × 1 0 1 4 . For a 6x6 board, k = 4 . 4 k = 4.4 k = 4 . 4 , there are 1 . 5 × 1 0 2 3 1.5 \times 10^{23} 1 . 5 × 1 0 2 3 nodes, and for a regular 8x8 chess board, k = 5 . 2 5 k = 5.25 k = 5 . 2 5 , there are 1 . 3 × 1 0 4 6 1.3 \times 10^{46} 1 . 3 × 1 0 4 6 . Of course, since there are multiple solutions to the problem we won’t have to explore every single node, but the fractional part of the nodes we do have to explore is just a constant multiplier which does not change the exponential nature of the problem. We will leave it as an exercise for you to see if you can express k k k as a function of the board size.

Luckily there is a way to speed up the eight-by-eight case so that it runs in under one second. In the code sample below we show the code that speeds up the traverse . This function, called warnsdorffs_heuristic when passed as the heuristic function to find_solution_for above will cause the next_vertices to be sorted prioritizing those who which have the fewest subsequent legal moves.

This may seem counterintutitive; why not select the node that has the most available moves? The problem with using the vertex with the most available moves as your next vertex on the path is that it tends to have the knight visit the middle squares early on in the tour. When this happens it is easy for the knight to get stranded on one side of the board where it cannot reach unvisited squares on the other side of the board. On the other hand, visiting the squares with the fewest available moves first pushes the knight to visit the squares around the edges of the board first. This ensures that the knight will visit the hard-to- reach corners early and can use the middle squares to hop across the board only when necessary. Utilizing this kind of knowledge to speed up an algorithm is called a heuristic. Humans use heuristics every day to help make decisions, heuristic searches are often used in the field of artificial intelligence. This particular heuristic is called Warnsdorff’s heuristic, named after H. C. Warnsdorff who published his idea in 1823, becoming the first person to describe a procedure to complete the knight’s tour.

For fun, here is a very large ( 1 3 0 × 1 3 0 130 \times 130 1 3 0 × 1 3 0 ) open knight’s tour created using Warnsdorff’s heuristic:

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Knight’s Tour Problem in C++

Knight’s Tour problem is a classic puzzle in which the goal is to move a knight on a chessboard such that the knight visits every square exactly once. The knight moves in an L-shape: two squares in one direction and then one square perpendicular to that or vice versa. In this article, we will learn how to solve this problem using the C++ language.

Consider an 8×8 chessboard, and the knight starts at position (0, 0). The goal is to find a sequence of moves that allows the knight to visit each square exactly once.

To solve the Knight’s Tour problem, there are two primary methods: Backtracking and Warnsdorff’s Rule. Here, we will discuss both methods using the C++ programming language.

Knight’s Tour using Backtracking in C++

A knight can move in an “L” shape, i.e., two squares in one direction and then one square perpendicular to that direction. Suppose the current position of the knight is (i,j). The possible moves are:

To solve this problem using backtracking, we try all possible moves for the knight and backtrack if we reach an invalid state. This method ensures that we explore all potential solutions.

Create an empty solution matrix to track the knight’s path. Place the knight at the starting position. Try all possible moves from the current position. If a move is valid, mark the move in the solution matrix. Recursively attempt to solve the tour from the new position. If the tour is complete, return the solution. If no valid moves are available, backtrack and try the next move.

C++ Program to Solve Knight’s Tour using Backtracking

The following C++ program demonstrates how we can solve the Knight’s tour problem using backtracking.

Time Complexity : O(8^(N^2)), where N is the size of the chessboard. This complexity arises because for each cell, we try up to 8 possible moves. Auxiliary Space: O(N^2), as we use a 2D array to store the knight’s path.

Knight’s Tour using Warnsdorff’s Algorithm in C++

Warnsdorff’s rule is a heuristic method to solve the Knight’s Tour problem. The rule states that the knight should always move to the square from which the knight will have the fewest onward moves. This minimizes the branching factor and reduces the likelihood of running into a dead end.

Set the initial position of the knight randomly on the board. Mark the board at the starting position with the move number “1”. For each subsequent move number from 2 to the total number of squares on the board: Determine all possible positions the knight can move to from the current position. Choose the position with the minimum number of onward moves. Mark the board at this new position with the current move number. Return the marked board, where each square is marked with the move number on which it was visited.

C++ Program to Solve Knight’s Tour using Warnsdorff’s Rule

The following C++ program demonstrates how we can solve the Knight’s tour problem using Warnsdorff’s algorithm.

Time Complexity: O(N^3), where N is the size of the chessboard. The complexity is reduced significantly due to the heuristic used. Auxiliary Space: O(N^2), as we use a 2D array to store the knight’s path.

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Knight's tour problem

What is knight's tour problem.

In the knight's tour problem, we are given with an empty chess board of size NxN, and a knight. In chess, the knight is a piece that looks exactly like a horse. Assume, it can start from any location on the board. Now, our task is to check whether the knight can visit all of the squares on the board or not. When it can visit all of the squares, then print the number of jumps needed to reach that location from the starting point.

There can be two ways in which a knight can finish its tour. In the first way, the knight moves one step and returns back to the starting position forming a loop which is called a closed tour . In the second way i.e. the open tour , it finishes anywhere in the board.

For a person who is not familiar with chess, note that the knight moves in a special manner. It can move either two squares horizontally and one square vertically or two squares vertically and one square horizontally in each direction. So, the complete movement looks like English letter 'L' .

Suppose the size of given chess board is 8 and the knight is at the top-left position on the board. The next possible moves are shown below −

Each cell in the above chess board holds a number, that indicates where to start and in how many moves the knight will reach a cell. The final values of the cell will be represented by the below matrix −

Remember, this problem can have multiple solutions, the above matrix is one possible solution.

One way to solve the knight's tour problem is by generating all the tours one by one and then checking if they satisfy the specified constraint or not. However, it is time consuming and not an efficient way.

Backtracking Approach to Solve Knight's tour problem

The other way to solve this problem is to use backtracking. It is a technique that tries different possibilities until a solution is found or all options are tried. It involves choosing a move, making it, and then recursively trying to solve the rest of the problem. If the current move leads to a dead end, we backtrack and undo the move, then try another one.

To solve the knight's tour problem using the backtracking approach, follow the steps given below −

Start from any cell on the board and mark it as visited by the knight.

Move the knight to a valid unvisited cell and mark it visited. From any cell, a knight can take a maximum of 8 moves.

If the current cell is not valid or not taking to the solution, then backtrack and try other possible moves that may lead to a solution.

Repeat this process until the moves of knight are equal to 8 x 8 = 64.

In the following example, we will practically demonstrate how to solve the knight's tour problem.

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8.1 Objectives
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8.8 Building the Word Ladder Graph
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8.11 The Knight’s Tour Problem
8.12 Building the Knight’s Tour Graph
8.13 Implementing Knight’s Tour
8.14 Knight’s Tour Analysis
8.15 General Depth First Search
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8.12. Building the Knight’s Tour Graph" data-toggle="tooltip">
8.14. Knight’s Tour Analysis' data-toggle="tooltip" >

8.13. Implementing Knight’s Tour ¶

The search algorithm we will use to solve the knight’s tour problem is called depth first search ( DFS ). Whereas the breadth first search algorithm discussed in the previous section builds a search tree one level at a time, a depth first search creates a search tree by exploring one branch of the tree as deeply as possible. In this section we will look at two algorithms that implement a depth first search. The first algorithm we will look at directly solves the knight’s tour problem by explicitly forbidding a node to be visited more than once. The second implementation is more general, but allows nodes to be visited more than once as the tree is constructed. The second version is used in subsequent sections to develop additional graph algorithms.

The knightTour function takes four parameters: n , the current depth in the search tree; path , a list of vertices visited up to this point; u , the vertex in the graph we wish to explore; and limit the number of nodes in the path. The knightTour function is recursive. When the knightTour function is called, it first checks the base case condition. If we have a path that contains 64 vertices, we return from knightTour with a status of True , indicating that we have found a successful tour. If the path is not long enough we continue to explore one level deeper by choosing a new vertex to explore and calling knightTour recursively for that vertex.

DFS also uses colors to keep track of which vertices in the graph have been visited. Unvisited vertices are colored white, and visited vertices are colored gray. If all neighbors of a particular vertex have been explored and we have not yet reached our goal length of 64 vertices, we have reached a dead end. When we reach a dead end we must backtrack. Backtracking happens when we return from knightTour with a status of False . In the breadth first search we used a queue to keep track of which vertex to visit next. Since depth first search is recursive, we are implicitly using a stack to help us with our backtracking. When we return from a call to knightTour with a status of False , in line 11, we remain inside the while loop and look at the next vertex in nbrList .

Let’s look at a simple example of knightTour in action. You can refer to the figures below to follow the steps of the search. For this example we will assume that the call to the getConnections method on line 6 orders the nodes in alphabetical order. We begin by calling knightTour(0,path,A,6)

knightTour starts with node A Figure 3 . The nodes adjacent to A are B and D. Since B is before D alphabetically, DFS selects B to expand next as shown in Figure 4 . Exploring B happens when knightTour is called recursively. B is adjacent to C and D, so knightTour elects to explore C next. However, as you can see in Figure 5 node C is a dead end with no adjacent white nodes. At this point we change the color of node C back to white. The call to knightTour returns a value of False . The return from the recursive call effectively backtracks the search to vertex B (see Figure 6 ). The next vertex on the list to explore is vertex D, so knightTour makes a recursive call moving to node D (see Figure 7 ). From vertex D on, knightTour can continue to make recursive calls until we get to node C again (see Figure 8 , Figure 9 , and Figure 10 ). However, this time when we get to node C the test n < limit fails so we know that we have exhausted all the nodes in the graph. At this point we can return True to indicate that we have made a successful tour of the graph. When we return the list, path has the values [A,B,D,E,F,C] , which is the order we need to traverse the graph to visit each node exactly once.

Figure 3: Start with node A ¶

Figure 4: Explore B ¶

Figure 5: Node C is a dead end ¶

Figure 6: Backtrack to B ¶

Figure 7: Explore D ¶

Figure 8: Explore E ¶

Figure 9: Explore F ¶

Figure 10: Finish ¶

Figure 11 shows you what a complete tour around an eight-by-eight board looks like. There are many possible tours; some are symmetric. With some modification you can make circular tours that start and end at the same square.

Figure 11: A Complete Tour of the Board ¶

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center(width ,fillchar)
count(string,begin,end)
endswith(suffix ,begin=0,end=len(string))
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find(substring ,beginIndex, endIndex)
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ljust(width[,fillchar])
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rjust(width,[,fillchar])
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split(str,num=string.count(str))
splitlines(num=string.count('\n'))
startswith(str,beg=0,end=len(str))
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zfill(width)
rpartition()

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The Knight's Tour Problem (using Backtracking Algorithm)
Knight Tour Problem Backtracking (Data Structures and Algorithms #8)(Recursion #5)(Backtracking #4)
8.12. Building the Knight’s Tour Graph
Figure 1 from A New Structured Knight Tour Algorithm by Four Knights
Knight Tour Problem and Its Graph Analysis
Solving Knight's Tour variation problem in python : r/learnprogramming

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This SotN Randomizer has INFINITE Complexity?!
The knight's tour problem #chess #vedantadesika
The Knight's Tour Problem #graphtheory #math #chess
Run Knight, Run!
Touring The Shard
Knight's Tour Problem(Backtracking)

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The Knight's tour problem
Time Complexity : There are N 2 Cells and for each, we have a maximum of 8 possible moves to choose from, so the worst running time is O(8 N^2). Auxiliary Space: O(N 2) Important Note: No order of the xMove, yMove is wrong, but they will affect the running time of the algorithm drastically. For example, think of the case where the 8th choice of the move is the correct one, and before that our ...
The Knight's Tour Problem (using Backtracking Algorithm)
The backtracking algorithm used to solve the Knight's Tour problem has an exponential time complexity. The number of possible paths for the knight grows very quickly as the size of the chessboard increases, which means that the time taken to explore all possible paths grows exponentially. The exact time complexity of the Knight's Tour ...
Warnsdorff's algorithm for Knight's tour problem
Warnsdorff's algorithm for Knight's tour problem. Problem : A knight is placed on the first block of an empty board and, moving according to the rules of chess, must visit each square exactly once. Following is an example path followed by Knight to cover all the cells. The below grid represents a chessboard with 8 x 8 cells.
Knight's Tour Problem in C
Time Complexity: O(8N^2), as there are N^2 Cells and for each, we have a maximum of 8 possible moves to choose from, so the worst running time is . Auxiliary Space: O(N^2) Knight's Tour using Warnsdorff's Algorithm in C. Warnsdorff's rule is a heuristic method to solve the Knight's Tour problem.
Knight's tour
Knight's graph showing all possible paths for a knight's tour on a standard 8 × 8 chessboard. The numbers on each node indicate the number of possible moves that can be made from that position. The knight's tour problem is an instance of the more general Hamiltonian path problem in graph theory.The problem of finding a closed knight's tour is similarly an instance of the Hamiltonian cycle ...
Knight's Tour Problem
The Knight's Tour Problem is one of the famous problem in which we have the knight on a chessboard. The knight is supposed to visit every square exactly once. We have explored Backtracking and Warnsdorff's algorithm. ... Time Complexity: O(n * n) for traversing the n x n squares.
PDF An efficient algorithm for the Knight's tour problem
The knight's tour problem is the problem of con- structing such a tour, given n. A knight ... Such a tour can be constructed in time 0(n2). Proof. The proof is by induction on n. The claim is easily seen to be true for 6 < n < 10 by inspecting Fig. 2 (the knight's tours in this figure were obtained using the random walk algorithm described ...
The Knight's Tour Problem in Java
Complexity Analysis: The time complexity of the Knight's tour problem using backtracking is O(8^(N^2)), where N is the size of the chessboard. It is because at each move, the Knight has 8 possible moves to choose from, and the total number of moves the Knight can make is N^2. The space complexity of the program is O(N^2).
8.14. Knight's Tour Analysis
The average branching factor is k = 3.8 So the number of nodes in the search tree is 3.8 25 − 1 or 3.12 × 10 14. For a 6x6 board, k = 4.4, there are 1.5 × 10 23 nodes, and for a regular 8x8 chess board, k = 5.25, there are 1.3 × 10 46. Of course, since there are multiple solutions to the problem we won't have to explore every single node ...
7.11. The Knight's Tour Problem
7.11. The Knight's Tour Problem ¶. Another classic problem that we can use to illustrate a second common graph algorithm is called the knight's tour. The knight's tour puzzle is played on a chess board with a single chess piece, the knight. The object of the puzzle is to find a sequence of moves that allow the knight to visit every ...
Backtracking Knight's Tour Problem
A knight's tour is a sequence of moves of a knight on a chessboard such that the knight visits every square only once. If the knight ends on a square that is one knight's move from the beginning square (so that it could tour the board again immediately, following the same path), the tour is closed, otherwise it is open.
PDF Knight Tour Problem and its Graph Analysis
Knight Graph are bipartite graph. In Knight graph, no two graph vertices within the same set are adjacent. A knight move always alternates between white and black squares. Knight graph are equivalent to two-colorable graph. Its chromatic number is 2. Knight graph is a special case of k-partite graph with k=2. Knight graph are perfect Graph.
8.11. The Knight's Tour Problem
The Knight's Tour Problem — Problem Solving with Algorithms and Data Structures. 8.11. The Knight's Tour Problem ¶. Another classic problem that we can use to illustrate a second common graph algorithm is called the "knight's tour.". The knight's tour puzzle is played on a chess board with a single chess piece, the knight.
An efficient algorithm for the Knight's tour problem
The algorithm of Theorem 2.1 is particularly easy to implement, and can be used to construct knight's tours of size up to 1000 x 1000 in under 11 s on a SUN SPARC 2. The construction described in the proof of Theorem 2.1 can also be used to obtain some knight's tours with interesting properties.
The Knight's tour problem
Following is the Backtracking algorithm for Knight's tour problem. If all squares are visited. print the solution. Else. a) Add one of the next moves to solution vector and recursively. check if this move leads to a solution. (A Knight can make maximum. eight moves. We choose one of the 8 moves in this step).
A Knight's Tour
The "knight's tour" is a classic problem in graph theory, first posed over 1,000 years ago and pondered by legendary mathematicians including Leonhard Euler before finally being solved in 1823. We will use the knight's tour problem to illustrate a second common graph algorithm called depth first search. The knight's tour puzzle is played on a chess board with a single chess piece ...
Knight's Tour Problem in C++
Knight's Tour problem is a classic puzzle in which the goal is to move a knight on a chessboard such that the knight visits every square exactly once. The knight moves in an L-shape: two squares in one direction and then one square perpendicular to that or vice versa. ... Time Complexity: O(8^(N^2)), where N is the size of the chessboard ...
9.11. The Knight's Tour Problem
9.11. The Knight's Tour Problem ¶. Another classic problem that we can use to illustrate a second common graph algorithm is called the "knight's tour.". The knight's tour puzzle is played on a chess board with a single chess piece, the knight. The object of the puzzle is to find a sequence of moves that allow the knight to visit ...
What is Knight's tour problem?
To solve the knight's tour problem using the backtracking approach, follow the steps given below −. Start from any cell on the board and mark it as visited by the knight. Move the knight to a valid unvisited cell and mark it visited. From any cell, a knight can take a maximum of 8 moves.
c++
The Knight's Tour problem can be stated as follows: Given a chess board with n × n squares, find a path for the knight that visits every square exactly once. ... I mean time complexity O(8^(n^2)) = k*8^(n^2) and the difference will be in a lower k factor. Algorithm improvements.
Similar problem to Knight's tour
Similar problem to Knight's tour. Ask Question Asked 8 months ago. Modified 8 months ago. Viewed 56 times 0 $\begingroup$ ... Time complexity of this solution to N-queens problem. 1. Filling a 3x3 board with connected tiles. 3. Maximize number of museums visited in a day. 6.
8.13. Implementing Knight's Tour
8.13. Implementing Knight's Tour¶. The search algorithm we will use to solve the knight's tour problem is called depth first search (DFS).Whereas the breadth first search algorithm discussed in the previous section builds a search tree one level at a time, a depth first search creates a search tree by exploring one branch of the tree as deeply as possible.
Knight Tour Problem
Knight Tour Problem. The Knight's Tour problem is a classic problem in the field of computational mathematics and computer science.It is a puzzle where a chess knight is placed on an empty chess board and the goal is to move the knight to every square on the board exactly once without re-visiting any squares. The tour is considered closed if the knight ends on the square it started on.