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On the different possible non-linear arrangements of eight men on a Chess-board

Published online by Cambridge University Press:  20 January 2009

T.B. Sprague
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The question having been proposed to me as a puzzle: To arrange eight men on a chess-board, so that no two of them shall be in the same line,—that is to say, that no two are to be in the same column, nor in the same row, nor in the same diagonal line,—I succeeded before very long in solving it by finding the annexed arrangement. (Fig. 45.)

Type
Research Article
Information
Proceedings of the Edinburgh Mathematical Society , Volume 8 , February 1889 , pp. 30 - 43
Copyright
Copyright © Edinburgh Mathematical Society 1889

References

* It is not necessary for our present purposes to investigate the laws according to which our symbols of operation, i, r, p, combine with each other; and I will therefore content myself with stating a few of the principal laws, without any demonstration:—

As an illustration of the use of these relations, let as take the processes by which arrangement (2) in Fig. 47 is got from No. (1); that is to say, the process of turning the chess board clock-wise through a quadrant. We have seen that (2)=ip(l), so that the operation will be denoted by ip. If, now, we repeat this process, we have (ip)2=ip(ip)=ip.pr=ip 2r=ir; and this, as we have seen, is the process by which (3) is got from (1). Again, if we repeat the same process once more, we have

and this, as we have seen, is the process by which (4) was got from (1).

As another illustration, I will show how, by the same operations i, r, p, all the aspects in the text, instead of being got from No. (1), are got from one of the others,—say (4).

It will be noticed that (rp)−1=pr; and similarly (ip)−1=pi, (irp)−1=pri=irp.

* This operation I denote by t (the initial letter of transpose), so that 46831752=t(24683175); whence (4A)=t(3a).

† This operation will be denoted by t −1, so that 52468317=t −1(24683175) and we have (4A)=t(3a)=t 2(8C).