A Tangled Tale

other two luncheons, on each assumption, as “one-and-ninepence” and “four-and-ten-pence” respectively, which harmony of results, he will say, “shows that the answers are correct.” And yet, as a matter of fact, the buns were 2 d. each, the queen-cakes 3 d., the sausage-rolls 6 d., and the Zoëdone 2 d. a bottle: so that Clara’s third luncheon had cost one-and-sevenpence, and her thirsty friends had spent four-and-fourpence!
Another remark of Balbus I will quote and discuss: for I think that it also may yield a moral for some of my readers. He says “it is the same thing in substance whether in solving this problem we use words and call it Arithmetic, or use letters and signs and call it Algebra.” Now this does not appear to me a correct description of the two methods: the Arithmetical method is that of “synthesis” only; it goes from one known fact to another, till it reaches its goal: whereas the Algebraical method is that of “analysis”: it begins with the goal, symbolically represented, and so goes backwards, dragging its veiled victim with it, till it has reached the full daylight of known facts, in which it can tear off the veil and say “I know you!”
Take an illustration. Your house has been broken into and robbed, and you appeal to the policeman who was on duty that night. “Well, Mum, I did see a chap getting out over your garden-wall: but I was a good bit off, so I didn’t chase him, like. I just cut down the short way to the Chequers, and who should I meet but Bill Sykes, coming full split round the corner. So I just ups and says ‘My lad, you’re wanted.’ That’s all I says. And he says ‘I’ll go along quiet, Bobby,’ he says, ‘without the darbies,’ he says.” There’s your Arithmetical policeman. Now try the other method. “I seed somebody a running, but he was well gone or ever I got nigh the place. So I just took a look round in the garden. And I noticed the footmarks, where the chap had come right across your flowerbeds. They was good big footmarks sure-ly. And I noticed as the left foot went down at the heel, ever so much deeper than the other. And I says to myself ‘The 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.





Hecla.


Martreb.






§1 (2 steps).


Adelaide.


Clifton  C.


 E. K. C.


Guy.


L’Inconnu.


Little Jack.


Nil desperandum.


Simple Susan.


Yellowhammer.


Woolly One.




§2 (3 steps).


 A. A.


A Christmas Carol.


Afternoon Tea.


An appreciative Schoolma’am.


Baby.


Balbus.


Bog-Oak.


The Red Queen.


Wallflower.




§3 (4 steps).


Hawthorn.


Joram.


 S. S. G.




§4 (5 steps).


A Stepney Coach.




§5 (6 steps).


Bay Laurel.


Bradshaw of the Future.




§6 (9 steps).


Old King Cole.




§7 (14 steps).


Theseus.








Answers to Correspondents
I have received several letters on the subjects of Knots II and VI, which lead me to think some further explanation desirable.
In Knot II, 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 VI, 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 further 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 VI 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,