Figure 6 - available via license: Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International
Content may be subject to copyright.
Source publication
The classic n-queens problem asks for placements of just n mutually non-attacking queens on an n × n board. By adding enough pawns, we can arrange to fill roughly one-quarter of the board with mutually non-attacking queens. How many pawns do we need? We discuss that question for square boards as well as rectangular m × n boards.
Contexts in source publication
Context 1
... example, on a 4 × 7 board we can place at most 4 2 7 2 = 23.5 = (2)(4) = 8 queens, as illustrated in Figure 6. If both m and n are odd, there is only one way to place the queens and every other square requires a pawn. ...
Context 2
... columns with labels of the form4k + 2, place pawns in the even-numbered rows and queens in the odd-numbered rows. We get a striped pattern as illustrated in Figure 6. We can check that none of the queens attack each other and that there are the right number of queens. ...
Context 3
... can check that none of the queens attack each other and that there are the right number of queens. We conclude by eliminating a pawn in the first and last non-empty columns, as indicated in Figure 6. If m or n is 2, that is the best possible result: A 2-row board without pawns can hold at most 2 queens, one per row. ...
Similar publications
The $n$-queens puzzle is to place $n$ mutually non-attacking queens on an $n \times n$ chess board. We present a simple randomized algorithm to construct such configurations. We first use a random greedy algorithm to construct an approximate toroidal $n$-queens configuration. In this well-known variant the diagonals wrap around the board from left...
Citations
... Conclusively, the problem has been told in the process of integer programming corresponding to the assignment program. The below figure [6] explains about one of the example for 8*8 queen problem. The solutions for n queen value in N*N chessboard are fixed. ...
The combinatorial optimization problem is a collection of problems which need a sample amount of time and effort to be solved. Vast difficulties have been occurring to solving these types of problem that there is no exact formula to solve the problem. Each feasible solution works on some order and the size of the probability increases algorithmically as the number of the problem also increases dynamically. This paper discusses about N-Queen problem, it is also a type of NP-hard problem. Many researchers have proposed various methods and algorithms for this problem. Henceforth, Genetic Algorithm is one kind of famous algorithm for solving NP hard problems. This paper mainly focuses on the review work of genetic algorithm to solve the N-Queen Problems (NPQ).
... Proof: On a 1 × 1 or a 2 × 2 board, a single dragon horse placed on any square leaves all other empty squares attacked, so β(H 1 ) = β(H 2 ) = 1. To see that β(H 4 ) 4, place dragon horses on squares (0, 0), (0, 2), (2, 1), and (2,3). To see that β(H 4 ) 4, partition the board into a 2 × 2 array of 2 × 2 blocks and note that each block can only hold one dragon horse of an independent set. ...
... 5. Since chess queens do not move through other pieces, placing pawns on a board may increase the maximum number of independent queens we can put on that board. In [3] it is noted that the maximum number of mutually nonattacking queens that can be placed on an n × n board with pawns is n 2 4 if n is even and (n+1) 2 ...
... That argument also works for dragon kings and dragon horses. So, as [3] asks for queens, how many pawns are needed to allow the maximum number of dragon kings or dragon horses on the board? ...
Given a (symmetrically-moving) piece from a chesslike game, such as shogi, and an n×n board, we can form a graph with a vertex for each square and an edge between two vertices if the piece can move from one vertex to the other. We consider two pieces from shogi: the dragon king, which moves like a rook and king from chess, and the dragon horse, which moves like a bishop and rook from chess. We show that the independence number for the dragon kings graph equals the independence number for the queens graph. We show that the (independent) domination number of the dragon kings graph is n − 2 for 4 ≤ n ≤ 6 and n − 3 for n ≥ 7. For the dragon horses graph, we show that the independence number is 2n − 3 for n ≥ 5, the domination number is at most n−1 for n ≥ 4, and the independent domination number is at most n for n ≥ 5.
