A Tangled Tale Part 14

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 a.s.sumed by most, if not all, of the compet.i.tors. TROJa.n.u.s however says "a.s.suming, 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 a.s.sumption?

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 rubbis.h.!.+" 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, 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.

To ill.u.s.trate this, let us take two candidates for a prize, _A_ and _B_, who are to compete in French, German, and Italian:

(_a_) Let it be laid down that the result is to depend on their _relative_ knowledge of each subject, so that, whether their marks, for French, be "1, 2" or "100, 200," the result will be the same: and let it also be laid down that, if they get equal marks on 2 papers, the final marks are to have the same ratio as those of the 3rd paper. This is a case of ordinary Double Rule of Three. We multiply _A_'s 3 marks together, and do the same for _B_. Note that, if _A_ gets a single "0,"

his final mark is "0," even if he gets full marks for 2 papers while _B_ gets only one mark for each paper. This of course would be very unfair on _A_, though a correct solution under the given conditions.

(_b_) The result is to depend, as before, on _relative_ knowledge; but French is to have twice as much weight as German or Italian. This is an unusual form of question. I should be inclined to say "the resulting ratio is to be nearer to the French ratio than if we multiplied as in (_a_), and so much nearer that it would be necessary to use the other multipliers _twice_ to produce the same result as in (_a_):" _e.g._ if the French Ratio were 9/10, and the others 4/9, 1/9 so that the ultimate ratio, by method (_a_), would be 2/45, I should multiply instead by 2/3, 1/3, giving the result, 1/3 which is nearer to 9/10 than if he had used method (_a_).

(_c_) The result is to depend on _actual_ amount of knowledge of the 3 subjects collectively. Here we have to ask two questions. (1) What is to be the "unit" (_i.e._ "standard to measure by") in each subject? (2) Are these units to be of equal, or unequal value? The usual "unit" is the knowledge shown by answering the whole paper correctly; calling this "100," all lower amounts are represented by numbers between "0" and "100." Then, if these units are to be of equal value, we simply add _A_'s 3 marks together, and do the same for _B_.

(_d_) The conditions are the same as (_c_), but French is to have double weight. Here we simply double the French marks, and add as before.

(_e_) French is to have such weight, that, if other marks be equal, the ultimate ratio is to be that of the French paper, so that a "0" in this would swamp the candidate: but the other two subjects are only to affect the result collectively, by the amount of knowledge shown, the two being reckoned of equal value. Here I should add _A_'s German and Italian marks together, and multiply by his French mark.

But I need not go on: the problem may evidently be set with many varying conditions, each requiring its own method of solution. The Problem in Knot VI. was meant to belong to variety (_a_), and to make this clear, I inserted the following pa.s.sage:

"Usually the compet.i.tors differ in one point only. Thus, last year, Fifi and Gogo made the same number of scarves in the trial week, and they were equally light; but Fifi's were twice as warm as Gogo's, and she was p.r.o.nounced twice as good."

What I have said will suffice, I hope, as an answer to BALBUS, who holds that (_a_) and (_c_) are the only possible varieties of the problem, and that to say "We cannot use addition, therefore we must be intended to use multiplication," is "no more illogical than, from knowledge that one was not born in the night, to infer that he was born in the daytime"; and also to FIFEE, who says "I think a little more consideration will show you that our 'error of _adding_ the proportional numbers together for each candidate instead of _multiplying_' is no error at all." Why, even if addition _had_ been the right method to use, not one of the writers (I speak from memory) showed any consciousness of the necessity of fixing a "unit" for each subject. "No error at all!" They were positively steeped in error!

One correspondent (I do not name him, as the communication is not quite friendly in tone) writes thus:--"I wish to add, very respectfully, that I think it would be in better taste if you were to abstain from the very trenchant expressions which you are accustomed to indulge in when criticising the answer. That such a tone must not be" ("be not"?) "agreeable to the persons concerned who have made mistakes may possibly have no great weight with you, but I hope you will feel that it would be as well not to employ it, _unless you are quite certain of being correct yourself_." The only instances the writer gives of the "trenchant expressions" are "hapless" and "malefactors." I beg to a.s.sure him (and any others who may need the a.s.surance: I trust there are none) that all such words have been used in jest, and with no idea that they could possibly annoy any one, and that I sincerely regret any annoyance I may have thus inadvertently given. May I hope that in future they will recognise the distinction between severe language used in sober earnest, and the "words of unmeant bitterness," which Coleridge has alluded to in that lovely pa.s.sage beginning "A little child, a limber elf"? If the writer will refer to that pa.s.sage, or to the preface to "Fire, Famine, and Slaughter," he will find the distinction, for which I plead, far better drawn out than I could hope to do in any words of mine.

The writer's insinuation that I care not how much annoyance I give to my readers I think it best to pa.s.s over in silence; but to his concluding remark I must entirely demur. I hold that to use language likely to annoy any of my correspondents would not be in the least justified by the plea that I was "quite certain of being correct." I trust that the knot-untiers and I are not on such terms as those!

I beg to thank _G. B._ for the offer of a puzzle--which, however, is too like the old one "Make four 9's into 100."

ANSWERS TO KNOT VIII.

-- 1. THE PIGS.

_Problem._--Place twenty-four pigs in four sties so that, as you go round and round, you may always find the number in each sty nearer to ten than the number in the last.

_Answer._--Place 8 pigs in the first sty, 10 in the second, nothing in the third, and 6 in the fourth: 10 is nearer ten than 8; nothing is nearer ten than 10; 6 is nearer ten than nothing; and 8 is nearer ten than 6.

This problem is noticed by only two correspondents. BALBUS says "it certainly cannot be solved mathematically, nor do I see how to solve it by any verbal quibble." NOLENS VOLENS makes Her Radiancy change the direction of going round; and even then is obliged to add "the pigs must be carried in front of her"!

-- 2. THE GRURMSTIPTHS.

_Problem._--Omnibuses start from a certain point, both ways, every 15 minutes. A traveller, starting on foot along with one of them, meets one in 12-1/2 minutes: when will he be overtaken by one?

_Answer._--In 6-1/4 minutes.

_Solution._--Let "_a_" be the distance an omnibus goes in 15 minutes, and "_x_" the distance from the starting-point to where the traveller is overtaken. Since the omnibus met is due at the starting-point in 2-1/2 minutes, it goes in that time as far as the traveller walks in 12-1/2; _i.e._ it goes 5 times as fast. Now the overtaking omnibus is "_a_"

behind the traveller when he starts, and therefore goes "_a_ + _x_"

while he goes "_x_." Hence _a_ + _x_ = 5_x_; _i.e._ 4_x_ = _a_, and _x_ = _a_/4. This distance would be traversed by an omnibus in 15/4 minutes, and therefore by the traveller in 5 15/4. Hence he is overtaken in 18-3/4 minutes after starting, _i.e._ in 6-1/4 minutes after meeting the omnibus.

Four answers have been received, of which two are wrong. DINAH MITE rightly states that the overtaking omnibus reached the point where they met the other omnibus 5 minutes after they left, but wrongly concludes that, going 5 times as fast, it would overtake them in another minute.

The travellers are 5-minutes-walk ahead of the omnibus, and must walk 1-4th of this distance farther before the omnibus overtakes them, which will be 1-5th of the distance traversed by the omnibus in the same time: this will require 1-1/4 minutes more. NOLENS VOLENS tries it by a process like "Achilles and the Tortoise." He rightly states that, when the overtaking omnibus leaves the gate, the travellers are 1-5th of "_a_" ahead, and that it will take the omnibus 3 minutes to traverse this distance; "during which time" the travellers, he tells us, go 1-15th of "_a_" (this should be 1-25th). The travellers being now 1-15th of "_a_" ahead, he concludes that the work remaining to be done is for the travellers to go 1-60th of "_a_," while the omnibus goes l-12th. The _principle_ is correct, and might have been applied earlier.

CLa.s.s LIST.

I.

BALBUS.

DELTA.

ANSWERS TO KNOT IX.

-- 1. THE BUCKETS.

_Problem._--Lardner states that a solid, immersed in a fluid, displaces an amount equal to itself in bulk. How can this be true of a small bucket floating in a larger one?