664 
Proceedings of the Royal Society 
b 
Whence P > 
the relation stated above (p. 653). 
Non-Inter sectors. 
As an example of the trigonometry of non-intersectors, I select the following 
formulae, the proof of which I leave to the reader. 
If KA and LB be two non-intersectors, K and L the points of least distance, 
KA = LB = r, KAB = LBA = <|>, KL =d, AB = D. 
Then sin A. — = sin h— cos h- . . . (13) 
2p 2p p 
cos h^- 
• ^ 2 p 
sm 0 = (14) 
eos h~ 
2 P 
The results of (6) to (11) are given by Newcomb (Borchardt, lxxxiii. p. 293) 
mostly without demonstration. He assumes formula (3) as one of the axioms on 
which he bases his synthesis. Although I have read most of the original 
literature on the subject, I am more immediately indebted to Newcomb and 
Frischauf for the materials of the foregoing sketch. 
2. Note on the Theory of the “15 Puzzle.” 
By Professor Tait. 
[After this note had been laid before the Council, the new 
number (vol. ii. No. 4) of the “American Journal of Mathematics” 
reached us. In it there are exhaustive papers by Messrs John- 
ston and Story on the subject of this American invention. The 
principles they give differ only in form of statement from those 
at which I had independently arrived. I have, therefore, cut 
down my paper to the smallest dimensions consistent with intelli- 
gibility.— P. G. T.] 
The essential feature of this puzzle is that the circulation of 
the pieces is necessarily in rectangular channels. Whether these 
form four-sided figures, or have any greater (even) number of sides, 
the number of squares in the channel itself is always even. (This 
is the same thing as saying that a rook’s re-entrant path always 
contains an even number of squares. This follows immediately 
from the fact that a rook always passes through black and white 
squares alternately. The same thing is true of a bishop’s re-entering 
