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



so that there is only one solution involving c 2 ; and as this only 

 involves c 2 , g 6 , b 8 , f 4 , and a 5 , the middle square on the board, 

 we obtain do fresh solution by the rotation of the board ; the 

 only other solution, therefore, is its reflexion, viz. g 4 b 2 a b c 8 f 6 . 

 There are thus ten solutions, which agree (after correcting a 

 couple of misprints) with those found by Dr. Giinther by deve- 

 loping the determinant (1) in full. 



It is convenient to distinguish the different classes of solutions 

 as ultimate, penultimate, &c, — the ultimate solutions being those 

 which involve no corner square, the penultimate those which, 

 not being ultimate, involve no square next to a corner square, 

 the antepenultimate those which, not being ultimate or penulti- 

 mate, involve no square next but one to a corner square, and so 

 on : thus for the board of 25 squares there are 8 ultimate, 2 

 penultimate, and no antepenultimate solutions. It is also con- 

 venient, in considering any class of solutions, to call the consti- 

 tuent which multiplies the minor which is actually developed the 

 leading square; and the three squares which would take its 

 place if the board were turned through 90°, 180°, and 270° vice 

 leading squares, or simply vice squares. Thus, for the board of 

 25 squares, c 2 is the leading penultimate square, and g 6 , b s ,f 4 

 are the vice squares. 



Consider now a board of 36 squares, 



96 



"l 



c 2 



e 3 



9t 



h 



«3 



C 4 



*s 



d* 



h 



«5 



% 



f* 



d, 



be 



«7 



h 



fe 



d 7 



*8 



m e 



h 



/e 



d 9 



'10 



'10 a \\ 



it will be noticed that every one of the 10 five-solutions involves 

 an a, so that there is no six-solution involving a n ; there are 

 therefore no ultimate six-solutions. On development we have 



d 3 a b . 



A b 6 a i 



h d 7 b 8 



• f 8 d 9 



9e 



H 



b w 



n 7 



9s 

 e Q 



+ e s 



d B b 4 

 f* d b 



fe h 



9e 



The first term gives, it will be found, only one penultimate solu- 

 tion involving c 2 , viz. c 2 e 5 ^ 8 / 4 ^ 7 6 10 , which involves the vice square 

 b l0 and is symmetrical : we thus by rotation obtain only one 

 other solution, viz. ^s/si^^io* These two solutions give by 



