chap's been a big hulking chap: and he goes lame on his left foot." And I rubs my hand on the wall where he got over, and there was soot on it, and no mistake. So I says to myself, "Now where can I light on a big man, in the chimbley-sweep line, what's lame of one foot?" And I flashes up permiscuous: and I says, "It's Bill Sykes!" says I.' There is your Algebraical policeman--a higher intellectual type, to my thinking, than the other.
LITTLE JACK's solution calls for a word of praise, as he has written out what really is an algebraical proof in words, without representing any of his facts as equations. If it is all his own, he will make a good algebraist in the time to come. I beg to thank SIMPLE SUSAN for some kind words of sympathy, to the same effect as those received from OLD CAT.
HECLA and MARTREB are the only two who have used a method certain either to produce the answer, or else to prove it impossible: so they must share between them the highest honours.
Class List
1. HECLA MARTREB
2. 1 (2 steps)
ADELAIDE CLIFTON C. E.K.C. GUY L'INCONNU LITTLE JACK
NIL DESPERANDUM SIMPLE
SUSAN YELLOW-HAMMER
WOOLLY ONE
2 (3 steps)
A. A. A CHRISTMAS CAROL AFTERNOON TEA
AN
APPRECIATIVE SCHOOL-MA'AM BABY BALBUS BOG-OAK
THE RED QUEEN WALL-FLOWER
3 (4
steps) 4 (5 steps)
HAWTHORN JORAM S.S.G. A STEPNEY COACH
5 (6 steps)
BAY LAUREL BRADSHAW
OF THE FUTURE
6 (9 steps) 7 (14 steps)
OLD KING COLE THESEUS
ANSWERS TO CORRESPONDENTS
I HAVE received several letters on the subjects of Knots 2 and 6, which lead me to think some further explanation desirable.
In Knot 2, I had intended the numbering of the houses to begin at one corner of the Square, and this was assumed by most, if not all, of the competitors. TROJANUS, however, says, `Assuming, in default of any information, that the street enters the square in the middle of each side, it may be supposed that the numbering begins at a street.' But surely the other is the more natural assumption?
In Knot 6, the first Problem was, of course a mere jeu de mots, whose presence I thought excusable in a series of Problems whose aim is to entertain rather than to instruct: but it has not escaped the contemptuous criticisms of two of my correspondents, who seem to think that Apollo is in duty bound to keep his bow always on the stretch. Neither of them has guessed it: and this is true human nature. Only the other day--the 31st of September, to be quite exact--I met my old friend Brown, and gave him a riddle I had just heard. With one great effort of his colossal mind, Brown guessed it. `Right!' said I. `Ah,' said he, `it's very neat--very neat. And it isn't an answer that would occur to everybody. Very neat indeed.' A few yards farther on, I fell in with Smith, and to him I propounded the same riddle. He frowned over it for a minute, and then gave it up. Meekly I faltered out the answer. `A poor thing, sir!' Smith growled, as he turned away. `A very poor thing! I wonder you care to repeat such rubbish!' Yet Smith's mind is, if possible, even more colossal than Brown's.
The second Problem of Knot 6 is an example in ordinary Double Rule of Three, whose essential feature is that the result depends on the variation of several elements, which are so related to it that, if all but one be constant, it varies as that one: hence, if none be constant, it varies as their product. Thus, for example, the cubical contents of a rectangular tank vary as its length, if breadth and depth be constant, and so on; hence, if none be constant, it varies as the product of the length, breadth, and depth.
When the result is not thus connected with the varying elements, the problem ceases to be Double Rule of Three and often becomes one of great complexity.