40 



♦ KNOWLEDGE ♦ 



[December 1, 1886. 



Tbere result mauy varieties of cutting, according to the Puzzle XIV. Shoir the same hij a method akin In the 



shape of the given and required rectangles. Figures •"), G, second. 



and 7 (Mogul's), show some of the varieties for the first 

 method {hi, in tig. 5, is equal to./'/(. Where either rectangle 

 is very long, more equidistant thwart cut.s parallel to .\F', 



or qf, will have to be made). Similar diversities will readUy 

 be dealt with in the case of the second method. 



PUZZLES FOR THIS MONTH. 

 Puzzle XIII. Sho^r hoic a jmmllelogram, «« abcd, fiy. 8, 

 may be divided by the fevest nvmber of lines so that the 



Fig 8 



Vig-! 



Fir 10 



pieces may Jit another par(dklo(/ram, ofkl, Jii/. 9, having 

 angle at a, equal to awjk at A, by a method akin to the first 

 described above. 



Puzzle XV. Divide abcd, fig. 8. into three pieces so as 

 to cover the parallelogram afkl,fiy. 10. 



THE WHIST-PLAYERS' PUZZLES. 



(See KxowLEDt:E for October and November.) 



T is required to arrange sixteen persons to play 

 whist on Jive days, each day retainitig their 

 jilaces unchanged throughout, so that no two 

 shall tv'ice sit at the same table. I take first 

 the subordinate puzzle : — To arranye twelve 

 persons to play whist on five days — single 

 dnvimy on four days, full whist only on the 

 remaining day — no tioo sitting twice at the same table. 

 The direct method of solution is similar to that employed 

 for the fifteen school-gii-ls. It would take little time to 

 run thiough the combinations, because wo might at once 

 set out three combinations of four each, the single day of 

 full whist, and these limit us effectively in making out 

 the sixteen combinations of three each out of which the 

 four days of single dummy are to be provided for. 

 But the simplest way of proceeding is as follows : — 

 Write after a four paiis of letters, and one triplet, in 

 alphabetical sequence, giving the first table of the following 

 set : — 



Table I. Table II. Table III. Table IV. Table V. 



To make the other sets of four take the two columns 

 headed b and c in the first table, getting the second table. 

 Get three sets beginning with b by taking e, o, and i from 

 the right-hand vertical column of Table I. with j, k, and l 

 of the lowest row ; and get three sets beginning with c in like 

 manner. These form Tables III. and IV. Lastly, get three 

 more combinations by taking d, e, and f, each with the only 

 pair of the letters g, h, i, j, k, l, remaining available — getting 

 Table V. 



From these sets we obtain quite easily the following 

 solution : — 



1st day. 2nd day. 3rd day. 4th day. 5th day. 



Take now the problem of the sixteen. We have fii-st to 

 get twenty combinations of four, no two in any of which 

 will be twice in the same set. To do this we begin with 

 Table I. belo\i', written out in the usual way : — 



We deal with the 

 (lined off in Table I 

 arrangements, just as 

 to the 2nd, ord, 4th, 



twelve lettors e f g . . . to p 



), so as to form from them four 



the four arrangements belonging 



and 5 th days above are formed 



