## Dancing links, algorithm X and the n-queens puzzle

January 20th, 2019 • Comments Off on Dancing links, algorithm X and the n-queens puzzleDonald Knuth’s 24th annual Christmas lecture inspired me to implement algorithm X using dancing links myself. To demonstrate how this algorithms works, I choose the n-queens puzzle. This problem has some interesting aspects to it which show all kinds of facets of the algorithm.

Solving the n-queens puzzle using algorithm X can be achieved as follows. Each queen being placed on a chess board will lead to 4 constraints that need to be cared for:

- There can only be one queen in the given row.
- There can only be one queen in the given column.
- There can only be one queen in the diagonal going from top left to bottom right.
- There can only be one queen in the diagonal going from bottom left to top right.

Let’s look at a chess board and determine some identifiers for each constraint that any placement can have (See the youtube video of the lecture at this timestep). Donald Knuth doesn’t go into too many details on how to solve this but let’s look at an 4×4 chessboard example:

Assigning an identifier can be done straight forward: column based constraints are numbered 0-3; row based constraints get the ids from 4-7; diagonals 8-12; reverse diagonals 13-17 for a chessboard of size 4. Note that you can ignore the diagonal of size 1, as they are automatically covered by the row or column the queen would be placed in.

Calculating the identifiers for the constraints is simple. Given an queen positioned at coordinates *(x, y)* on a chessboard of size *N* these are the formulations you can use:

- the row constraint equals
*x* - the column constraint equals
*y + N* - for the diagonal constraint equals
*x + y + (2N-1)* - for the reverse diagonal constraint equals
*(N – 1 – x + y) + (4N-4)*

Note that each constraint identifier calculation we use an offset, so we can all represent them in one context: 0 for the row constraints, N for the column constraints, 2N-1 for the diagonal and 4N-4 for the reverse diagonal constraints. For the diagonal constraints you need to ensure they are within their ranges: *2*N-1 < constraint < 3*N* and *3*N+1 < constraint < 4*N + 1*. If the calculated identifier of the constraints is not within the ranges, you can just drop the constraint – this is the case for queens being placed at corner positions. Placing a queen at coordinates *(0,0)* will result in 3 constraints: *[0, 4, 15]*. Placing a queen at coordinates *(1,2)* will results in 4 constraints: *[1, 6, 10, 16]*.

Now that for each queen that can be placed we can calculate the uniquely identifiable constraints identifier, we can translate this problem into an exact cover problem. We are looking for a set of placed queens, for which all constraints can be satisfied. Let’s look at a matrix representation (for a chessboard of size 4, in which the rows are individual possible placements options of a queen, while the columns represent the constraints. The 1s indicate that it is a constraint that needs to be satisfied for the given placement.

Queen coordinates | Primary constraints | Secondary constraints | ||||||||||||||||

x-coord | y-coord | diagonals | r-diagonals | |||||||||||||||

0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | |

0,0 | 1 | 1 | 1 | |||||||||||||||

0,1 | 1 | 1 | 1 | 1 | ||||||||||||||

0,2 | 1 | 1 | 1 | 1 | ||||||||||||||

0,3 | 1 | 1 | 1 | |||||||||||||||

1,0 | 1 | 1 | 1 | 1 | ||||||||||||||

1,1 | 1 | 1 | 1 | 1 | ||||||||||||||

1,2 | 1 | 1 | 1 | 1 | ||||||||||||||

1,3 | 1 | 1 | 1 | 1 | ||||||||||||||

2,0 | 1 | 1 | 1 | 1 | ||||||||||||||

2,1 | 1 | 1 | 1 | 1 | ||||||||||||||

2,2 | 1 | 1 | 1 | 1 | ||||||||||||||

2,3 | 1 | 1 | 1 | 1 | ||||||||||||||

3,0 | 1 | 1 | 1 | 1 | ||||||||||||||

3,1 | 1 | 1 | 1 | 1 | ||||||||||||||

3,2 | 1 | 1 | 1 | 1 | ||||||||||||||

3,3 | 1 | 1 | 1 |

Note that the table columns have been grouped into two categories: primary and secondary constraints. This is also a lesser known aspect of the algorithms at hand – primary columns and their constraints **must be covered exactly once**, while the secondary columns **must only be covered at most once**. Looking at the n-queens puzzle, this make sense – to fill the chessboard to the maximum possible number of queens, each row and column should have a queen, while there might be diagonals which are not blocked by a queen.

Implementing algorithm x using dancing links is reasonable simple – just follow the instructions in Knuth’s paper here. Let’s start with some Go code for the data structures:

type x struct { left *x right *x up *x down *x col *column data string } type column struct { x size int name string } type Matrix struct { head *x headers []column solutions []*x }

Note the little change I made to store a data string in the object x structure. This will be useful when printing the solutions later on. The following function in the can be used to initialize a new matrix:

func NewMatrix(n_primary, n_secondary int) *Matrix { n_cols := n_primary + n_secondary matrix := &Matrix{} matrix.headers = make([]column, n_cols) matrix.solutions = make([]*x, 0, 0) matrix.head = &x{} matrix.head.left = &matrix.headers[n_primary-1].x matrix.head.left.right = matrix.head prev := matrix.head for i := 0; i < n_cols; i++ { column := &matrix.headers[i] column.col = column column.up = &column.x column.down = &column.x if i < n_primary { // these are must have constraints. column.left = prev prev.right = &column.x prev = &column.x } else { // as these constraints do not need to be covered - do not weave them in. column.left = &column.x column.right = &column.x } } return matrix }

Now that we can instantiate the matrix data structure according to Knuth's description the actual search algorithm can be run. The *search* function can roughly be split into 3 parts: a) if we found a solution - print it, b) finding the next column to cover and c) backtrack search through the remaining columns:

func (matrix *Matrix) Search(k int) bool { j := matrix.head.right if j == matrix.head { // print the solution. fmt.Println("Solution:") for u := 0; u < len(matrix.solutions); u += 1 { fmt.Println(u, matrix.solutions[u].data) } return true } // find column with least items. column := j.col for j = j.right; j != matrix.head; j = j.right { if j.col.size < column.size { column = j.col } } // the actual backtrack search. cover(column) matrix.solutions = append(matrix.solutions, nil) for r := column.down; r != &column.x; r = r.down { matrix.solutions[k] = r for j = r.right; j != r; j = j.right { cover(j.col) } matrix.Search(k + 1) for j = r.left; j != r; j = j.left { uncover(j.col) } } matrix.solutions = matrix.solutions[:k] uncover(column) return false }

Note that when printing the possible solution(s) we use the *data* field, described earlier for the *x object*. The *cover* and *uncover* functions needed for the *search* function to work are pretty straight forward:

func cover(column *column) { column.right.left = column.left column.left.right = column.right for i := column.down; i != &column.x; i = i.down { for j := i.right; j != i; j = j.right { j.down.up = j.up j.up.down = j.down j.col.size-- } } } func uncover(column *column) { for i := column.up; i != &column.x; i = i.up { for j := i.left; j != i; j = j.left { j.col.size++ j.down.up = j j.up.down = j } } column.right.left = &column.x column.left.right = &column.x }

Now there is only only little helper functions missing, providing an easy way to add the constraints. Note the data field of the *x object* data structure is set here as well:

func (matrix *Matrix) AddConstraints(constraints []int, name string) { new_x_s := make([]x, len(constraints)) last := &new_x_s[len(constraints)-1] for i, col_id := range constraints { // find the column matching the constraint. column := &matrix.headers[col_id] column.size++ // new node x. x := &new_x_s[i] x.up = column.up x.down = &column.x x.left = last x.col = column x.up.down = x x.down.up = x x.data = name last.right = x last = x } }

Given an list of constraints to add - for example for the queen at position *(3,2): [3, 6, 12, 14]* - this little helper function will add the new x objects to the matrix and weave them in

This is basically all there is to tell about this algorithm. Now you can easily solve your (exact) cover problems. Just create a new matrix, add the constraints and run the search algorithm. Doing so for a 4x4 chess board queens placement problem the code above will give you:

Solution: 0 01 1 13 2 20 3 32 Solution: 0 02 1 10 2 23 3 31

In future it would be nice how dancing links could be used to solve other e.g. graph search algorithms, such as A* or similar.better other ways to implement it.