458 Mr. J. W. L. Glaisher on the Problem of the Eight Queens. 



interesting account of the history of the problem is given in the 

 last Number of Grunert's Archiv der Mathematik und Physik by 

 Dr. Siegmund Gunther*, who has also suggested a new way of 

 solving the question, which it is the object of this communica- 

 tion to develop. 



Dr. Gunther (considering a board of n 2 squares, and of course 

 n queens) remarks that if the determinant 



a \ c <2 



e 3 



94 



h • • 



K a 3 



<>4 



*5 



96 • • 



d 3 b 4 



#5 



C 6 



e 7 . . 



h ^ 



h 



a 7 



c s . . 



h fe 



d 7 



h 



a 9 . . 



°2n-2 a 2n 



be expanded, and all the terms be rejected in which either the 

 same letter or the same suffix appears more than once, then the 

 terms that remain will give all the solutions of the problem. 

 The reason for the rule is evident : from the nature of a deter- 

 minant each term involves one constituent from each row and 

 one from each column, and the terms thus represent all the po- 

 sitions in which the queens cannot take one another castle- 

 fashion ; the omission of the terms in which the same letter or 

 suffix appears more than once excludes the cases in which two 

 or more queens lie on the same diagonal (i. e. can take one? 

 another bishop-fashion), so that the terms that remain are the 

 solutions. Dr. Gunther develops the determinants for boards 

 of 9, 16, and 25 squares^ but, owing to the number of terms 

 involved, does not proceed further; he remarks that for the 

 chessboard of 64 squares it would be necessary to calculate 

 20,160 terms. 



Of course it would be quite out of the question to actually 

 write down twenty thousand terms ; but in the way which will 

 now be explained, I found it a lengthy piece of work cer- 

 tainly, but still not a very laborious task, to find by means of Dr. 

 Gunther's principle all the solutions for boards of 36, 49, and 64 

 squares. 



Starting from the beginning, it is easily seen that on a board 

 of 9 squares there is no solution, and that on a board of 16 

 squares there are two solutions, viz. c 2 e 5 d 3 b 6 and e 3 b 2 c 6 d 5 . To 

 find the solutions for a board of 25 squares, 



* Zur mathematischen Theorie des Schachbretts, vol. lvi. part 3, pp. 



281-292. 



