Chess Board Puzzles: How Many Queens Fit, With and Without Pawns

How many queens can share a chessboard without any of them attacking another? Under the classic rules the answer is one queen per row, so eight on a standard board ^[1]. Add pawns to block the queens' lines and the number climbs: a single pawn already makes room for nine queens, and with enough pawns about a quarter of the board can hold queens ^[1].

This article follows that question and its relatives through peer-reviewed papers: the classic queens puzzle, what pawns change, a piece called the dragon king, and a newer question about how far apart two squares can be.

What is the classic queens puzzle?

The puzzle asks you to place queens on a chessboard, at most one per square, so that no queen attacks another ^[1]. A queen attacks any square it could reach in one move along its row, its column or a diagonal ^[1].

The best-known version uses eight queens on the standard board. It was first proposed in 1848, and it was later widened to n queens on a board with n rows and n columns ^[1]. The wider version has a solution on a board of side 1 and on every board of side 4 or more ^[1]. On boards of side 2 and 3 there is no solution at all ^[1].

Why can there never be more queens than rows? Each queen attacks every other square in its own row and its own column, so a row can hold only one queen, and an n by n board can hold at most n queens ^[1].

You can try the smallest interesting case yourself. On a four by four board, put a queen on the second square of the top row, the fourth square of the next row, the first square of the third row and the third square of the bottom row. No two of them share a row, a column or a diagonal.

What changes when pawns join the board?

In chess, a queen cannot move through another piece ^[1]. So if a pawn stands between two queens on the same line, those queens no longer attack each other ^[1]. That simple rule opens up a new puzzle: with some pawns on the board, how many queens can you fit?

This is called the maximum queens problem, and it was first posed in a 1998 doctoral dissertation ^[1]. Some answers are now known:

• On any board of side 6 or more, one pawn is enough to fit one extra queen ^[1].
• On the standard board, that means nine queens with one pawn ^[1].
• To fit 6 queens on a board of side 5, you need 3 pawns, as that dissertation proved ^[1].
• For any number k of pawns, k extra queens fit once the board is large enough ^[1].

The pattern is easy to state, but the details are not. Small boards behave differently from large ones, and the exact size at which each pattern starts to work is part of what makes the puzzle interesting.

What is the most queens any board can hold?

If you may use as many pawns as you like, there is still a ceiling. Divide the board into blocks of two by two squares. Any two queens in the same block sit on neighbouring squares, so they attack each other no matter where the pawns go ^[1]. A board can therefore never hold more queens than it has blocks: n²/4 on a board of even side n, and (n + 1)²/4 on a board of odd side ^[1].

That ceiling can be reached. Put a queen on every other square of every other row, and fill every remaining square with a pawn ^[1]. Each queen is then fenced off from all the others. On the standard board, the ceiling works out at 16 queens, a quarter of the 64 squares ^[1].

The open question is how few pawns you need to reach that ceiling. A 2016 paper discussed it for square boards and for rectangular boards of m rows and n columns ^[1].

What is the n + k queens problem?

A closely related puzzle fixes the numbers in advance. On an n by n board you place k pawns and n + k queens, so that any two queens on the same row, column or diagonal have at least one pawn between them ^[2].

For 1, 2 or 3 pawns, a solution exists on every board whose side is at least 5 more than the number of pawns ^[2]. For 4 or more pawns, solutions are known on boards whose side is more than 25 times the number of pawns, but researchers suspect the true lower limit is far smaller than that ^[2].

A counting argument explains why the numbers fit together so neatly. A row with no pawns can hold only one piece, and each pawn added to a row raises its capacity by at most one ^[2]. So k pawns allow at most n + k dragon kings or queens on the whole board ^[2]. To reach that total, every stretch of every row and column between pawns must be used: pieces and pawns alternate, and each row and column starts and ends with a piece ^[2].

Another striking fact concerns symmetry. On any board of side greater than 1, no arrangement of k pawns and n + k queens looks the same after being reflected in a line ^[2]. Dragon kings, as the next section shows, behave differently.

What is a dragon king?

The dragon king comes from shogi, a board game related to chess. It moves any number of squares along a row or a column, like a rook, or one square diagonally, like a king ^[2]. It cannot move through or jump over other pieces ^[2].

A 2018 paper replaced the queens in the n + k problem with dragon kings ^[2]. The question becomes: can you place k pawns and n + k dragon kings on an n by n board so that no dragon king is one move away from another ^[2]? The paper showed that a solution exists whenever the board's side is at least 5 more than the number of pawns ^[2].

The paper also looked at symmetry. It built solutions that look the same after a quarter turn or a half turn, solutions that match their reflection in one or both diagonals, and solutions with no symmetry at all ^[2]. Given some extra conditions, solutions in each symmetry class exist on boards whose side is at least twice the number of pawns plus 5 ^[2].

How do these puzzles connect to matrices?

There is a neat bridge from the board to a grid of numbers. Write 1 for each dragon king, −1 for each pawn and 0 for each empty square ^[2]. In a full solution, the result is an alternating sign matrix: a grid of 0, 1 and −1 in which the nonzero numbers alternate in sign along every row and column, and the first and last nonzero number in each row and column is 1 ^[2]. Solutions to the dragon kings problem match exactly those alternating sign matrices in which no two 1s touch diagonally ^[2]. A similar link had already been found for the n + k queens problem ^[2].

Bridges like this let mathematicians move a question from one field to another, where different tools may solve it.

How far apart can two squares be?

A newer question turns the board into a map of routes. On an empty board, a single rook or queen can reach any square in two moves ^[3]. What happens to that number when pawns stand in the way?

To study this, a 2021 paper treated the board as a graph: each free square is a point, and two points are joined when a piece can move between them in one move ^[3]. The distance between two squares is the fewest moves needed, and the diameter of the board is the largest distance between any two squares ^[3]. The paper asked how many pawns are needed so that the diameter grows to a chosen number ^[3].

The answers so far, for boards of side 4 or more, are tidy ^[3]:

• One pawn is enough to make some trip need 3 moves, for a rook or for a queen ^[3].
• For a rook, a pawn in any square except the four corners does this, while a pawn in a corner changes nothing ^[3].
• Two pawns are needed, and enough, to make some rook trip take 4 moves ^[3].
• Three pawns are needed, and enough, to make some rook trip take 5 or 6 moves, or some queen trip take 4 moves ^[3].

The same paper listed open questions, including what happens with unlimited pawns, on rectangular or cylinder-shaped boards, and with other pieces such as the knight ^[3].

Why do mathematicians keep coming back to chessboards?

Chessboard puzzles are quick to state and easy to check, yet they lead to deep questions. Hundreds of papers have been written on the queens problem and its variations alone, including versions on cylinders, ring-shaped boards and three-dimensional boards, and versions with bishops, rooks and invented pieces ^[1].

They also link up with other parts of mathematics. The dragon kings problem connects to alternating sign matrices ^[2], and the distance question is really a question about graphs ^[3]. A puzzle that fits on a board you can hold turns out to touch ideas used far beyond it.

Try it yourself

Here are three small challenges that use the ideas above. None needs a real board: a grid drawn on paper is enough.

• Place five queens on a five by five board so that none attacks another. Hint: two queens in neighbouring rows can never sit in neighbouring columns. One answer, reading down the rows: the first, third, fifth, second and fourth squares.
• Use the block argument above to show why a four by four board can never hold more than four queens, however many pawns you add.
• On an empty board, a rook needs at most two moves to reach any square. Where would you put one pawn so that some trip needs a third move?

If you enjoy reasoning about squares, routes and blocked lines, the riddles below use the same kind of thinking.