Chapter 6

"Aftur Kyng Aruirag, of wam we habbeth y told, Marius ys sone was kyng, quoynte mon & bold. And ys sone was aftur hym, _Coil_ was ys name, Bothe it were quoynte men, & of noble fame."

BALBUS lays it down as a general principle that "in order to ascertain the cost of any one luncheon, it must come to the same amount upon two different assumptions." (_Query._ Should not "it" be "we"? Otherwise the _luncheon_ is represented as wishing to ascertain its own cost!) He then makes two assumptions--one, that sandwiches cost nothing; the other, that biscuits cost nothing, (either arrangement would lead to the shop being inconveniently crowded!)--and brings out the unknown luncheons as 8_d._ and 19_d._, on each assumption. He then concludes that this agreement of results "shows that the answers are correct." Now I propose to disprove his general law by simply giving _one_ instance of its failing. One instance is quite enough. In logical language, in order to disprove a "universal affirmative," it is enough to prove its contradictory, which is a "particular negative." (I must pause for a digression on Logic, and especially on Ladies' Logic. The universal affirmative "everybody says he's a duck" is crushed instantly by proving the particular negative "Peter says he's a goose," which is equivalent to "Peter does _not_ say he's a duck." And the universal negative "nobody calls on her" is well met by the particular affirmative "_I_ called yesterday." In short, either of two contradictories disproves the other: and the moral is that, since a particular proposition is much more easily proved than a universal one, it is the wisest course, in arguing with a Lady, to limit one's _own_ assertions to "particulars," and leave _her_ to prove the "universal" contradictory, if she can. You will thus generally secure a _logical_ victory: a _practical_ victory is not to be hoped for, since she can always fall back upon the crushing remark "_that_ has nothing to do with it!"--a move for which Man has not yet discovered any satisfactory answer. Now let us return to BALBUS.) Here is my "particular negative," on which to test his rule. Suppose the two recorded luncheons to have been "2 buns, one queen-cake, 2 sausage-rolls, and a bottle of Zoëdone: total, one-and-ninepence," and "one bun, 2 queen-cakes, a sausage-roll, and a bottle of Zoëdone: total, one-and-fourpence." And suppose Clara's unknown luncheon to have been "3 buns, one queen-cake, one sausage-roll, and 2 bottles of Zoëdone:" while the two little sisters had been indulging in "8 buns, 4 queen-cakes, 2 sausage-rolls, and 6 bottles of Zoëdone." (Poor souls, how thirsty they must have been!) If BALBUS will kindly try this by his principle of "two assumptions," first assuming that a bun is 1_d._ and a queen-cake 2_d._, and then that a bun is 3_d._ and a queen-cake 3_d._, he will bring out the other two luncheons, on each assumption, as "one-and-nine-pence" 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 foot-marks, where the chap had come right across your flower-beds. They was good big foot-marks 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.

I.

HECLA. MARTREB.

II.

§ 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 SCHOOLMA'AM. BABY. BALBUS. BOG-OAK. THE RED QUEEN. WALL-FLOWER.

§ 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, 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 illustrate 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 passage:

"Usually the competitors 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 pronounced 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 assure him (and any others who may need the assurance: 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 passage beginning "A little child, a limber elf"? If the writer will refer to that passage, 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 pass 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 1-12th. The _principle_ is correct, and might have been applied earlier.

CLASS 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?

_Solution._--Lardner means, by "displaces," "occupies a space which might be filled with water without any change in the surroundings." If the portion of the floating bucket, which is above the water, could be annihilated, and the rest of it transformed into water, the surrounding water would not change its position: which agrees with Lardner's statement.

* * * * *

Five answers have been received, none of which explains the difficulty arising from the well-known fact that a floating body is the same weight as the displaced fluid. HECLA says that "only that portion of the smaller bucket which descends below the original level of the water can be properly said to be immersed, and only an equal bulk of water is displaced." Hence, according to HECLA, a solid, whose weight was equal to that of an equal bulk of water, would not float till the whole of it was below "the original level" of the water: but, as a matter of fact, it would float as soon as it was all under water. MAGPIE says the fallacy is "the assumption that one body can displace another from a place where it isn't," and that Lardner's assertion is incorrect, except when the containing vessel "was originally full to the brim." But the question of floating depends on the present state of things, not on past history. OLD KING COLE takes the same view as HECLA. TYMPANUM and VINDEX assume that "displaced" means "raised above its original level," and merely explain how it comes to pass that the water, so raised, is less in bulk than the immersed portion of bucket, and thus land themselves--or rather set themselves floating--in the same boat as HECLA.

I regret that there is no Class-list to publish for this Problem.

* * * * *

§ 2. BALBUS' ESSAY.

_Problem._--Balbus states that if a certain solid be immersed in a certain vessel of water, the water will rise through a series of distances, two inches, one inch, half an inch, &c., which series has no end. He concludes that the water will rise without limit. Is this true?

_Solution._--No. This series can never reach 4 inches, since, however many terms we take, we are always short of 4 inches by an amount equal to the last term taken.

* * * * *

Three answers have been received--but only two seem to me worthy of honours.

TYMPANUM says that the statement about the stick "is merely a blind, to which the old answer may well be applied, _solvitur ambulando_, or rather _mergendo_." I trust TYMPANUM will not test this in his own person, by taking the place of the man in Balbus' Essay! He would infallibly be drowned.

OLD KING COLE rightly points out that the series, 2, 1, &c., is a decreasing Geometrical Progression: while VINDEX rightly identifies the fallacy as that of "Achilles and the Tortoise."

CLASS LIST.

I.

OLD KING COLE. VINDEX.

* * * * *

§ 3. THE GARDEN.

_Problem._--An oblong garden, half a yard longer than wide, consists entirely of a gravel-walk, spirally arranged, a yard wide and 3,630 yards long. Find the dimensions of the garden.

_Answer._--60, 60-1/2.

_Solution._--The number of yards and fractions of a yard traversed in walking along a straight piece of walk, is evidently the same as the number of square-yards and fractions of a square-yard, contained in that piece of walk: and the distance, traversed in passing through a square-yard at a corner, is evidently a yard. Hence the area of the garden is 3,630 square-yards: _i.e._, if _x_ be the width, _x_ (_x_ + 1/2) = 3,630. Solving this Quadratic, we find _x_ = 60. Hence the dimensions are 60, 60-1/2.

* * * * *

Twelve answers have been received--seven right and five wrong.

C. G. L., NABOB, OLD CROW, and TYMPANUM assume that the number of yards in the length of the path is equal to the number of square-yards in the garden. This is true, but should have been proved. But each is guilty of darker deeds. C. G. L.'s "working" consists of dividing 3,630 by 60. Whence came this divisor, oh Segiel? Divination? Or was it a dream? I fear this solution is worth nothing. OLD CROW'S is shorter, and so (if possible) worth rather less. He says the answer "is at once seen to be 60 × 60-1/2"! NABOB'S calculation is short, but "as rich as a Nabob" in error. He says that the square root of 3,630, multiplied by 2, equals the length plus the breadth. That is 60.25 × 2 = 120-1/2. His first assertion is only true of a _square_ garden. His second is irrelevant, since 60.25 is _not_ the square-root of 3,630! Nay, Bob, this will _not_ do! TYMPANUM says that, by extracting the square-root of 3,630, we get 60 yards with a remainder of 30/60, or half-a-yard, which we add so as to make the oblong 60 × 60-1/2. This is very terrible: but worse remains behind. TYMPANUM proceeds thus:--"But why should there be the half-yard at all? Because without it there would be no space at all for flowers. By means of it, we find reserved in the very centre a small plot of ground, two yards long by half-a-yard wide, the only space not occupied by walk." But Balbus expressly said that the walk "used up the whole of the area." Oh, TYMPANUM! My tympa is exhausted: my brain is num! I can say no more.

HECLA indulges, again and again, in that most fatal of all habits in computation--the making _two_ mistakes which cancel each other. She takes _x_ as the width of the garden, in yards, and _x_ + 1/2 as its length, and makes her first "coil" the sum of _x_-1/2, _x_-1/2, _x_-1, _x_-1, _i.e._ 4_x_-3: but the fourth term should be _x_-1-1/2, so that her first coil is 1/2 a yard too long. Her second coil is the sum of _x_-2-1/2, _x_-2-1/2, _x_-3, _x_-3: here the first term should be _x_-2 and the last _x_-3-1/2: these two mistakes cancel, and this coil is therefore right. And the same thing is true of every other coil but the last, which needs an extra half-yard to reach the _end_ of the path: and this exactly balances the mistake in the first coil. Thus the sum total of the coils comes right though the working is all wrong.