Brute-force algorithms

A brute-force search for a knight's tour is impractical on all but the smallest boards.^[18] For example, there are approximately 4×10⁵¹ possible move sequences on an 8 × 8 board,^[19] and it is well beyond the capacity of modern computers (or networks of computers) to perform operations on such a large set. However, the size of this number is not indicative of the difficulty of the problem, which can be solved "by using human insight and ingenuity ... without much difficulty."^[18]

Divide-and-conquer algorithms

By dividing the board into smaller pieces, constructing tours on each piece, and patching the pieces together, one can construct tours on most rectangular boards in linear time – that is, in a time proportional to the number of squares on the board.^[11][20]

Warnsdorf's rule

	a	b	c	d	e	f	g	h
8									8
7									7
6									6
5									5
4									4
3									3
2									2
1									1
	a	b	c	d	e	f	g	h

A graphical representation of Warnsdorf's Rule. Each square contains an integer giving the number of moves that the knight could make from that square. In this case, the rule tells us to move to the square with the smallest integer in it, namely 2.

A very large (130 × 130) square open knight's tour created using Warnsdorf's Rule

Warnsdorf's rule is a heuristic for finding a single knight's tour. The knight is moved so that it always proceeds to the square from which the knight will have the fewest onward moves. When calculating the number of onward moves for each candidate square, we do not count moves that revisit any square already visited. It is possible to have two or more choices for which the number of onward moves is equal; there are various methods for breaking such ties, including one devised by Pohl^[21] and another by Squirrel and Cull.^[22]

This rule may also more generally be applied to any graph. In graph-theoretic terms, each move is made to the adjacent vertex with the least degree.^[23] Although the Hamiltonian path problem is NP-hard in general, on many graphs that occur in practice this heuristic is able to successfully locate a solution in linear time.^[21] The knight's tour is such a special case.^[24]

The heuristic was first described in "Des Rösselsprungs einfachste und allgemeinste Lösung" by H. C. von Warnsdorf in 1823.^[24]

A computer program that finds a knight's tour for any starting position using Warnsdorf's rule was written by Gordon Horsington and published in 1984 in the book Century/Acorn User Book of Computer Puzzles.^[25]

Neural network solutions

Closed knight's tour on a 24 × 24 board solved by a neural network

The knight's tour problem also lends itself to being solved by a neural network implementation.^[26] The network is set up such that every legal knight's move is represented by a neuron, and each neuron is initialized randomly to be either "active" or "inactive" (output of 1 or 0), with 1 implying that the neuron is part of the solution. Each neuron also has a state function (described below) which is initialized to 0.

When the network is allowed to run, each neuron can change its state and output based on the states and outputs of its neighbors (those exactly one knight's move away) according to the following transition rules:

$${\displaystyle U_{t+1}(N_{i,j})=U_{t}(N_{i,j})+2-\sum _{N\in G(N_{i,j})}V_{t}(N)}$$

$${\displaystyle V_{t+1}(N_{i,j})=\left\{{\begin{array}{ll}1&{\mbox{if}}\,\,U_{t+1}(N_{i,j})>3\\0&{\mbox{if}}\,\,U_{t+1}(N_{i,j})<0\\V_{t}(N_{i,j})&{\mbox{otherwise}},\end{array}}\right.}$$

where ${\displaystyle t}$ represents discrete intervals of time, ${\displaystyle U(N_{i,j})}$ is the state of the neuron connecting square ${\displaystyle i}$ to square ${\displaystyle j}$, ${\displaystyle V(N_{i,j})}$ is the output of the neuron from ${\displaystyle i}$ to ${\displaystyle j}$, and ${\displaystyle G(N_{i,j})}$ is the set of neighbors of the neuron.

Although divergent cases are possible, the network should eventually converge, which occurs when no neuron changes its state from time ${\displaystyle t}$ to ${\displaystyle t+1}$. When the network converges, either the network encodes a knight's tour or a series of two or more independent circuits within the same board.