Arranging pieces on the chessboard, or other sizes and shapes of board, to fulfil various conditions, gives rise to many interesting problems. Those that use pieces of only one type are more of a mathematical nature, whereas problems with multiple types of piece approach the domain of true chess problems.
Arrangements of n marks on a board n×n so that there is one mark in each rank and file are termed satins and there are n! such arrays, since the mark in the first file can be placed in n ways, then the mark in the second file in n–1 ways and so on. The question of how many geometrically distinct satins T(n) there are of size n was considered by Kraitchik in his column in L'Echiquier (1925) and later in his book on Mathematical Recreations (1953 p.245). This work was completed by D. Holt in his article “Rooks Inviolate” Mathematical Gazette (1974 pp.131-4). The sequence runs: 1, 1, 2, 7, 23, 115, 694, 5282, 46066, 456454, ... The chart below shows all the geometrically distinct satins of sizes 1 to 5.
The term satin comes from weaving, where the files represent the warp threads and the ranks represent the weft threads and the marks indicate where the weft passes under the warp. Mathematicians will recognise that if the marks are 1s and vacant cells are 0s the array is a permutation matrix (which when used to multiply another array, under the usual definition of matrix multiplication, permutes its ranks or files). If we take the marks to represent chess rooks on a board, then satins solve the problem of placing as many rooks on the board as possible with no rook guarding or attacking any other. They also solve the problem of putting as few rooks on the board as possible so that all cells are either occupied or guarded, but not both.
A theorem, attributed variously to W. Burnside or G. Polya, states that the number of geometrically distinct patterns of a given type in a shape unaltered by a Group of k symmetry operations S1, S2, ..., Sk is (N1 + N2 + ... + Nk)/k, where Nr is the number of cases unaltered by operation Sr. One of these operations, which we may take to be S1, must be the identity operation; thus N1 is the total number of cases without regard to symmetry.
In the case of a square pattern there are 8 symmetry operations: S1–S4 rotation through 0, 1, 2 and 3 right angles (clockwise); and S5–S8 equal to S1–S4 combined with transposition (i.e. reflection in the principal diagonal, top left to bottom right). The 1×1 satin is the only one unaltered by reflection in the horizontal S6 or vertical S8; for all other cases (n>1) we have N6 = N8 = 0. We always have N2 = N4, cases unaltered by 90 degree rotation clockwise or anticlockwise, and N5 = N7, cases unaltered by reflection in principal or secondary diagonal.
Thus the formula becomes T(n) = [n! + 2N2(n) + N3(n) + 2N5(n)]/8 for n>1. Holt showed that N2(n) = N2(n–1) + (n–1)N2(n–2) and Kraitchik showed that N3(2k+1) = N3(2k) = 2kN3(2k–2) and N5(4k) = N5(4k+1) = (4k–2)N5(4k–4), N5(4k+2) = N5(4k+3) = 0. These recurrence relations enable us to calculate successive values of T(n).
The problem of arranging eight chess queens on a chessboard so that none guards or attacks another was apparently first proposed by Max Bezzel in the Berliner Schachzeitung 1848 and first fully solved by Franz Nauck in Leipziger Illustrierte Zeitung 1850. Since queens act along ranks, files or diagonals the problem is equivalent to choosing eight cells, one in each rank and file and no two in any diagonal. There are twelve geometricaly distinct solutions, one of which is symmetric; shown here.
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If we count rotations and reflections of an arrangement as different solutions then the 11 asymmetric solutions each appear in 8 forms and the symmetric solution in 4, giving the total 92 which is sometimes cited. We also take no account of the chequering of the board, which would double the count to 184, half with the top left corner light and half with it dark.
The eight officers are the pieces that make up each player's back row in the chess opening position, i.e. two Rooks, two Knights, two Bishops, Queen and King. The problem of arranging the eight White officers to guard all 64 squares of the chessboard can be solved, but only by putting the two bishops on the same coloured squares (when the board is chequered) as shown in the first diagram below — but this is usually considered to be cheating. If the bishops have to be on opposite coloured squares the best that can be achieved is 63 squares guarded. The unguarded square can be occupied by any piece except the King.
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