Yekimov Sergey
Publishing House “Education and
Science” , s.r.o.
rusnauka@email.cz
Praha , Czech republic
A particular solution
of the problem of n queens.
The method of successive
finding of cones of independent queens.
Key words: n independent queens , chessboard of size n x n ,
problem P=NP
1. Formulation of the problem.
The task of arranging n independent queens (so they do not threaten each other) on an empty chessboard of size n × n is known for a long time and has a simple solution for any chessboard size of 2m, where m is any integer greater than 1. Consider a more complicated task, there is a chessboard of size n x n.
On this board there is a Q area with already installed queens. It is necessary to place the maximum number of independent queens on the rest of the board so that they do not threaten each other and they do not threaten the queens who are located in the area of Q.
One of the variants of solving this problem is, in practice, using the method of enumerating positions when the computer program generates combinations of positions of the queens, and then the criterion of their independence is checked. The disadvantage of this option, according to the author, is a sharp increase in the amount of computation in solving problems with a large chess board size. With the increase in the size of the chessboard, the number of possible combinations increases dramatically and the computer can not cope with the volume of computations produced and "hangs".
In this paper, we
consider a different method, which the author calls the method of sequential finding of the coordinates of independent
queens.
The basis of this method is the postulate:
Each subsequent
queen is placed on a chessboard in such a place that, after its placement, the
number of cells of the chessboard not being under attack is minimal.
2 . Solution.
Suppose there is a chess board N of
size n × n on which there is a region Q with located queens, and there is
an area M of the chessboard fields that are not under the impact of the queens
located in the area Q. Let the number of fields that are in M equal m. Let's make a
table T of size m x m. Each cell of this table T corresponds to a strictly
defined cell on the chessboard N. We fill the table as follows. If, on appearing on the
chessboard N in the area M, the queen on the cell under the number i will threaten
the fields j that have the numbers {n1, n2, ..., nj).
In the cells (i, n1), (i, n2), ..., (i, nj),
(n1, i), (n2, i), ..., (nj, i), and also (i,
i) we set the digit 1. This operation is performed for all cells from the
region M. We leave the remaining cells empty.
The resulting matrix has the
following property.
If we sum up the numbers in the column or row
under the number i, we get the number of fields from the area M that are
compromised when the queen is placed in the cell numbered i , while the author of the field on
which the queen is located also considers himself to be at his stroke.
The summation over the terms and
columns gives the same result. For simplicity, we shall carry out the summation
over the columns. Summing the data across all the columns, we find those
columns whose sum of numbers is minimal. Suppose we have found the only such
column, and it has the number k. Then we delete from row T a row under number
k, column number k, all the lines that contain the digit 1 at the intersection
with column k, and all the columns that contain the digit 1 at the intersection
with the string k. Next, we put on the chessboard N in the cell corresponding
to the number k of the queen. Then the operations are repeated. Summarize the
numbers of Table T without taking into account the cells that were crossed out.
We find a column with a minimal sum, suppose it has the number g. We delete
from the table T a row under the number g, a column under the number g, all the
columns that at the intersection with the line at the number g have the number
1 and all the lines that at the intersection with the column under the number g
have the number 1. We put on the chessboard N of the queen, which corresponds
to the number g. These operations are repeated until it is possible. For large boards , the
number of columns that have the minimum amount can be several. Performing
operations with each of these columns, we obtain a separate solution of the
problem. Individual solutions may coincide.
As examples, consider a solution for
a 16 x 16 board, on which there are two areas
with previously installed queens, they are indicated by the red letter W, Fig.1 .Cells
under the firing of queens are indicated in yellow.
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Fig.1 The starting position, the
fields under attack and the numbering of the impact-free fields.
The cells free from impact turned out 36, numbered
them and we compile the table T, in our case, it has a size of 36 x 36.
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Fig. 2 Table T.
We summarize the values of the table T by rows
and columns. The sums for rows and columns are shaded in yellow.
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1 |
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3 |
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1 |
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1 |
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4 |
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1 |
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15 |
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5 |
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15 |
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6 |
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1 |
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1 |
1 |
1 |
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1 |
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1 |
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1 |
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15 |
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7 |
1 |
1 |
1 |
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9 |
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8 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
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1 |
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1 |
1 |
1 |
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1 |
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16 |
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9 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
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1 |
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1 |
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1 |
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15 |
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10 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
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1 |
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1 |
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1 |
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1 |
16 |
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11 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
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1 |
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1 |
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1 |
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1 |
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16 |
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12 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
1 |
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1 |
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1 |
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1 |
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1 |
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17 |
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13 |
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1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
1 |
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1 |
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1 |
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1 |
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15 |
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14 |
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1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
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1 |
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1 |
1 |
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1 |
1 |
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15 |
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15 |
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1 |
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1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
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1 |
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1 |
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1 |
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14 |
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16 |
1 |
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1 |
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1 |
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1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
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1 |
15 |
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17 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
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1 |
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1 |
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16 |
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18 |
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1 |
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1 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
1 |
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1 |
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15 |
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19 |
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1 |
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1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
1 |
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1 |
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15 |
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20 |
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1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
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1 |
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1 |
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13 |
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21 |
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1 |
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1 |
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1 |
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1 |
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1 |
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1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
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1 |
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1 |
15 |
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22 |
1 |
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1 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
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15 |
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23 |
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1 |
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1 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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16 |
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24 |
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1 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
1 |
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15 |
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25 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
1 |
1 |
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|
14 |
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26 |
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1 |
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1 |
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1 |
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1 |
1 |
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1 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
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|
14 |
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|
27 |
1 |
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1 |
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1 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
|
1 |
1 |
1 |
1 |
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|
13 |
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|
28 |
|
1 |
1 |
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1 |
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1 |
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1 |
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1 |
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1 |
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|
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
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|
15 |
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|
29 |
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1 |
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1 |
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1 |
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1 |
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|
1 |
1 |
1 |
1 |
1 |
1 |
1 |
|
1 |
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|
12 |
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|
30 |
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1 |
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1 |
1 |
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1 |
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1 |
1 |
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1 |
1 |
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1 |
9 |
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31 |
1 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
1 |
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1 |
9 |
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32 |
1 |
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1 |
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1 |
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1 |
1 |
1 |
1 |
1 |
|
8 |
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|
33 |
|
1 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
1 |
1 |
1 |
10 |
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|
34 |
1 |
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1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
9 |
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|
35 |
|
1 |
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1 |
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1 |
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1 |
1 |
1 |
1 |
1 |
8 |
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36 |
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1 |
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1 |
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1 |
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1 |
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1 |
1 |
|
1 |
1 |
1 |
1 |
10 |
|
|
Σ |
15 |
16 |
15 |
15 |
15 |
15 |
9 |
16 |
15 |
16 |
16 |
17 |
15 |
15 |
14 |
15 |
16 |
15 |
15 |
13 |
15 |
15 |
16 |
15 |
14 |
14 |
13 |
15 |
12 |
9 |
9 |
8 |
10 |
9 |
8 |
10 |
Fig. 3 Summarizing the T table by rows and columns.
The minimum value of the sum equal to 8 corresponds to the columns and
rows under the numbers 35 and 32. For simplicity, consider the row and column
at number 35. We cross out the line number 35, column number 35, the lines that
at the intersection with column number 35 have the value 1, which at the
intersection with line number 35 have the value 1. The crossed out area is
filled with a blue color.
|
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
13 |
14 |
15 |
16 |
17 |
18 |
19 |
20 |
21 |
22 |
23 |
24 |
25 |
26 |
27 |
28 |
29 |
30 |
31 |
32 |
33 |
34 |
35 |
36 |
Σ |
|
|
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
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1 |
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1 |
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1 |
1 |
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1 |
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15 |
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|
2 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
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1 |
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1 |
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1 |
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1 |
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1 |
|
16 |
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|
3 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
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1 |
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1 |
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1 |
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1 |
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1 |
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1 |
15 |
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|
4 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
1 |
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1 |
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1 |
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1 |
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1 |
1 |
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15 |
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5 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
1 |
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1 |
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1 |
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1 |
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1 |
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1 |
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15 |
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|
6 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
1 |
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1 |
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1 |
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1 |
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1 |
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1 |
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15 |
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|
7 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
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9 |
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|
8 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
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1 |
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1 |
1 |
1 |
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1 |
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16 |
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|
9 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
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1 |
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1 |
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1 |
|
15 |
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|
10 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
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1 |
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1 |
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1 |
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1 |
16 |
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|
11 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
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1 |
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1 |
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1 |
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1 |
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16 |
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12 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
1 |
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1 |
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1 |
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1 |
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1 |
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17 |
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13 |
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1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
1 |
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1 |
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1 |
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1 |
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15 |
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14 |
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1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
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1 |
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1 |
1 |
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1 |
1 |
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15 |
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15 |
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1 |
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1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
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1 |
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1 |
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1 |
|
14 |
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16 |
1 |
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1 |
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1 |
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1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
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1 |
15 |
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17 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
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1 |
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1 |
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16 |
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18 |
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1 |
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1 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
1 |
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1 |
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15 |
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19 |
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1 |
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1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
1 |
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1 |
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15 |
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20 |
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1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
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1 |
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1 |
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13 |
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21 |
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1 |
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1 |
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1 |
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1 |
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1 |
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1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
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1 |
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1 |
15 |
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22 |
1 |
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1 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
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15 |
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23 |
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1 |
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1 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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16 |
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24 |
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1 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
1 |
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15 |
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25 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
|
|
1 |
1 |
1 |
1 |
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|
14 |
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|
26 |
|
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1 |
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1 |
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1 |
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1 |
1 |
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1 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
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|
14 |
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|
27 |
1 |
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1 |
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1 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
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13 |
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|
28 |
|
1 |
1 |
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1 |
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1 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
|
1 |
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|
15 |
|
|
29 |
|
|
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Σ |
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16 |
15 |
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15 |
15 |
9 |
16 |
15 |
16 |
16 |
17 |
15 |
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14 |
15 |
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15 |
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13 |
15 |
15 |
16 |
15 |
14 |
14 |
13 |
15 |
12 |
9 |
9 |
8 |
10 |
9 |
8 |
10 |
Fig.4 Finding a place to install the first queen.
Thus, we found
the location of the first queen, this is cell number 35, on our board this
corresponds to the field (12, 2).
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0 |
1 |
2 |
3 |
4 |
5 |
6 |
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7 |
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11 |
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13 |
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W |
W |
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14 |
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W |
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15 |
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W |
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16 |
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W |
W |
W |
W |
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|
Fig. 5 Position
after the installation of the first queen.
Make sure that the queen located on the field (12, 2) does not come into conflict with the previously installed queens. We take the next step, summarize the remaining rows and columns after deletion. The minimum value of the sum equal to 5 corresponds to the column and the line under the number 31. Strike out the line number 31, column number 31, the lines that at the intersection with column number 31 have the value 1, and the columns that at the intersection with line number 31 have the value 1.
The crossed out area is red. Thus, we found the location of the second queen, this is cell number 31, on our board this field (10, 1). We establish the second queen, on this cell Fig. 7, and make sure that he does not conflict with the previously installed queens.
|
0 |
1 |
2 |
3 |
4 |
5 |
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24 |
25 |
26 |
27 |
28 |
29 |
30 |
31 |
32 |
33 |
34 |
35 |
36 |
Σ |
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1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
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13 |
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4 |
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1 |
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1 |
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14 |
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5 |
1 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
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1 |
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1 |
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1 |
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1 |
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14 |
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6 |
1 |
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14 |
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7 |
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1 |
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1 |
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1 |
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1 |
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9 |
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10 |
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1 |
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1 |
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1 |
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1 |
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1 |
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1 |
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13 |
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11 |
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1 |
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1 |
1 |
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1 |
1 |
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1 |
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1 |
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15 |
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12 |
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1 |
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1 |
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1 |
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1 |
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1 |
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1 |
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16 |
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13 |
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1 |
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1 |
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1 |
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1 |
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12 |
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17 |
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1 |
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1 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
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1 |
1 |
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1 |
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1 |
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14 |
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18 |
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1 |
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1 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
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1 |
1 |
1 |
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1 |
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14 |
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19 |
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1 |
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1 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
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1 |
1 |
1 |
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1 |
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14 |
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20 |
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1 |
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1 |
1 |
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1 |
1 |
1 |
1 |
1 |
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1 |
1 |
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1 |
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1 |
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12 |
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21 |
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1 |
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1 |
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1 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
1 |
1 |
1 |
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1 |
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13 |
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22 |
1 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
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14 |
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23 |
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1 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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15 |
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24 |
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1 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
1 |
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15 |
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25 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
1 |
1 |
1 |
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14 |
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26 |
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1 |
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1 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
1 |
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13 |
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27 |
1 |
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1 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
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1 |
1 |
1 |
1 |
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12 |
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28 |
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1 |
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1 |
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1 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
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1 |
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14 |
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29 |
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1 |
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1 |
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1 |
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1 |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
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11 |
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30 |
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1 |
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1 |
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1 |
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1 |
1 |
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1 |
1 |
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7 |
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31 |
1 |
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1 |
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1 |
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1 |
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1 |
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5 |
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32 |
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0 |
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33 |
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0 |
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34 |
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35 |
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36 |
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0 |
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Σ |
11 |
0 |
13 |
14 |
14 |
14 |
6 |
11 |
0 |
13 |
15 |
16 |
14 |
14 |
0 |
12 |
14 |
14 |
14 |
12 |
13 |
14 |
15 |
15 |
14 |
13 |
12 |
14 |
11 |
7 |
5 |
0 |
0 |
0 |
0 |
0 |
Fig. 6 Finding the
location for the second queen.
|
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
13 |
14 |
15 |
16 |
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1 |
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6 |
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W |
W |
W |
W |
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7 |
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W |
W |
W |
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8 |
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9 |
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10 |
W |
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11 |
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12 |
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W |
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Fig. 7 Position after the installation of the second queen.
We take the next step, summarize the remaining rows and columns after deletion. The minimum value of the sum equal to 5 corresponds to the columns and lines under the numbers 7 and 30. For simplicity, consider the row and column number 30. Strikethrough line 30, column number 30, lines that are at the intersection with column number 30 have a value of 1, and columns that are at the intersection with line number 30 have a value of 1. The crossed out area is filled with brown. Thus, we found the location of the second queen, this is cell number 30, on our board this field (8, 7). We establish the third queen, on this cell Fig. 9, we are convinced that he does not conflict with the queens established earlier.
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Fig. 8 Finding the location for the third queen.
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Fig. 9 Position after the installation of the third queen. We take the next step, summarize the remaining rows and columns after deletion. The minimum value of the sum equal to 5 corresponds to the columns and lines under the number 7. Strikethrough line number 7, column number 7, lines that are at the intersection with column number 7 have a value of 1, and columns that are at the intersection with line number 7 have a value of 1. The crossed out area is shaded in green. Thus, we found the location of the fourth queen, this is cell number 7, on our board this field (1, 12). We fix the fourth queen, on this cell Fig. 11 and make sure that it does not conflict with the previously installed queens. 9 Position after the installation of the third queen.
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27 |
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0 |
Fig. 10 Finding a place for installing the queen. |
0 |
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Fig.11 Position after installation of the fourth queen.
We take the next step, summarize the remaining rows and columns after deletion. The minimum value of the sum equal to 7 corresponds to the columns and lines under the number 10, 27. For simplicity, consider the row and column at number 10. Strike line number 10, column number 10, lines that are at the intersection with column number 10 have a value of 1, and the columns that at the intersection with line number 10 have a value of 1. The crossed out area is filled with purple color. Thus, we found the location of the fifth queen, this is cell number 10, on our board this corresponds to the field (2, 3). We fix the fifth queen, on this cell Fig. 13, make sure that he does not conflict with the previously installed queens.|
0 |
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11 |
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19 |
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23 |
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9 |
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25 |
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27 |
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7 |
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28 |
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29 |
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Fig. 12 Finding
the location for the fifth queen.
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Fig. 13 Position
after the installation of the fifth queen.
We take the next step , summarize the
remaining rows and columns after deletion. The minimum value of the sum equal
to 4 corresponds to the columns and lines under the numbers 18, 22, 27. For
simplicity, consider the row and column at number 18. We cross out the line
number 18, column number 18, the lines that at the intersection with column
number 18 have the value 1, and the columns that at the intersection with line
number 18 have the value 1.
The crossed out area is
darkened in blue. Thus, we found the location of the sixth queen, this is cell
number 18, on our board this field (3, 5). We fix the sixth queen, on this cell
in Fig. 15, we are convinced that it does not conflict with the queens
established earlier
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27 |
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1 |
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28 |
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Σ |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
5 |
4 |
5 |
0 |
4 |
0 |
6 |
0 |
0 |
4 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
Fig. 14 Finding the location for the sixth queen|
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
13 |
14 |
15 |
16 |
|
1 |
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6 |
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W |
W |
W |
W |
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7 |
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W |
W |
W |
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8 |
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9 |
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11 |
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13 |
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W |
W |
W |
W |
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14 |
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W |
W |
W |
W |
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15 |
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W |
W |
W |
W |
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16 |
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W |
W |
W |
W |
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Fig. 15
3. Conclusions.
Thus, we managed to establish six
additional queens. This method is applicable to chessboards at least up to the
size of 214 cells and can be implemented on Microsoft Excel. The
algorithm analyzes a tree of variants, each branch of which corresponds to a
separate solution.
4.
Literature
1.
Ian P. Gent, Christopher Jefferson and Peter Nightingale (2017)
"Complexity of n-Queens Completion", Volume 59, pages 815-848 doi:10.1613/jair.5512