Complete Works of Lewis Carroll

by Lewis Carroll

  I.

  B. E. D.

  C. H.

  Constance Johnson.

  Greystead.

  Guy.

  Hoopoe.

  J. F. A.

  M. A. H.

  Number Five.

  Pedro.

  R. E. X.

  Seven Old Men.

  Vis Inertiæ.

  Willy B.

  Yahoo.

  II.

  American Subscriber.

  An appreciative schoolma'am.

  Ayr.

  Bradshaw of the Future.

  Cheam.

  C. M. G.

  Dinah Mite.

  Duckwing.

  E. C. M.

  E. N. Lowry.

  Era.

  Euroclydon.

  F. H. W.

  Fifee.

  G. E. B.

  Harlequin.

  Hawthorn.

  Hough Green.

  J. A. B.

  Jack Tar.

  J. B. B.

  Kgovjni.

  Land Lubber.

  L. D.

  Magpie.

  Mary.

  Mhruxi.

  Minnie.

  Money-Spinner.

  Nairam.

  Old Cat.

  Polichinelle.

  Simple Susan.

  S. S. G.

  Thisbe.

  Verena.

  Wamba.

  Wolfe.

  Wykehamicus.

  Y. M. A. H.

  III.

  A. C. M.

  Arvon Marlborough Boy.

  Chilpome.

  Dublin Boy.

  Furze-Bush.

  James.

  Partridge.

  R. W.

  Waiting for the Train.

  ANSWERS TO KNOT V.

  Problem.—To mark pictures, giving 3 x's to 2 or 3, 2 to 4 or 5, and 1 to 9 or 10; also giving 3 o's to 1 or 2, 2 to 3 or 4 and 1 to 8 or 9; so as to mark the smallest possible number of pictures, and to give them the largest possible number of marks.

  Answer.—10 pictures; 29 marks; arranged thus:—

  x

  x

  x

  x

  x

  x

  x

  x

  x

  o

  x

  x

  x

  x

  x

  o

  o

  o

  o

  x

  x

  o

  o

  o

  o

  o

  o

  o

  o

  Solution.—By giving all the x's possible, putting into brackets the optional ones, we get 10 pictures marked thus:—

  x

  x

  x

  x

  x

  x

  x

  x

  x

  (x)

  x

  x

  x

  x

  (x)

  x

  x

  (x)

  By then assigning o's in the same way, beginning at the other end, we get 9 pictures marked thus:—

  (o)

  o

  (o)

  o

  o

  o

  (o)

  o

  o

  o

  o

  o

  o

  o

  o

  All we have now to do is to run these two wedges as close together as they will go, so as to get the minimum number of pictures——erasing optional marks where by so doing we can run them closer, but otherwise letting them stand. There are 10 necessary marks in the 1st row, and in the 3rd; but only 7 in the 2nd. Hence we erase all optional marks in the 1st and 3rd rows, but let them stand in the 2nd.

  Twenty-two answers have been received. Of these 11 give no working; so, in accordance with what I announced in my last review of answers, I leave them unnamed, merely mentioning that 5 are right and 6 wrong.

  Of the eleven answers with which some working is supplied, 3 are wrong. C. H. begins with the rash assertion that under the given conditions "the sum is impossible. For," he or she adds (these initialed correspondents are dismally vague beings to deal with: perhaps "it" would be a better pronoun), "10 is the least possible number of pictures" (granted): "therefore we must either give 2 x's to 6, or 2 o's to 5." Why "must," oh alphabetical phantom? It is nowhere ordained that every picture "must" have 3 marks! Fifee sends a folio page of solution, which deserved a better fate: she offers 3 answers, in each of which 10 pictures are marked, with 30 marks; in one she gives 2 x's to 6 pictures; in another to 7; in the 3rd she gives 2 o's to 5; thus in every case ignoring the conditions. (I pause to remark that the condition "2 x's to 4 or 5 pictures" can only mean "either to 4 or else to 5": if, as one competitor holds, it might mean any number not less than 4, the words "or 5" would be superfluous.) I. E. A. (I am happy to say that none of these bloodless phantoms appear this time in the class-list. Is it IDEA with the "D" left out?) gives 2 x's to 6 pictures. She then takes me to task for using the word "ought" instead of "nought." No doubt, to one who thus rebels against the rules laid down for her guidance, the word must be distasteful. But does not I. E. A. remember the parallel case of "adder"? That creature was originally "a nadder": then the two words took to bandying the poor "n" backwards and forwards like a shuttlecock, the final state of the game being "an adder." May not "a nought" have similarly become "an ought"? Anyhow, "oughts and crosses" is a very old game. I don't think I ever heard it called "noughts and crosses."

  In the following Class-list, I hope the solitary occupant of III. will sheathe her claws when she hears how narrow an escape she has had of not being named at all. Her account of the process by which she got the answer is so meagre that, like the nursery tale of "Jack-a-Minory" (I trust I. E. A. will be merciful to the spelling), it is scarcely to be distinguished from "zero."

  CLASS LIST.

  I.

  Guy.

  Old Cat.

  Sea-Breeze.

  II.

  Ayr.

  Bradshaw of the Future.

  F. Lee.

  H. Vernon.

  III.

  Cat.

  ANSWERS TO KNOT VI.

  Problem 1.—A and B began the year with only 1,000l. a-piece. They borrowed nought; they stole nought. On the next New-Year's Day they had 60,000l. between them. How did they do it?

  Solution.—They went that day to the Bank of England. A stood in front of it, while B went round and stood behind it.

  Two answers have been received, both worthy of much honour. Addlepate makes them borrow "0" and steal "0," and uses both cyphers by putting them at the right-hand end of the 1,000l., thus producing 100,000l., which is well over the mark. But (or to express it in Latin) At Spes infracta has solved it even more ingeniously: with the first cypher she turns the "1" of the 1,000l. into a "9," and adds the result to the original sum, thus getting 10,000l.: and in this, by means of the other "0," she turns the "1" into a "6," thus hitting the exact 60,000l.

  CLASS LIST

  I.

  At Spes Infracta.

  II.

  Addlepate.

  Problem 2.—L makes 5 scarves, while M makes 2: Z makes 4 while L makes 3. Five scarves of Z's weigh one of L's; 5 of M's weigh 3 of Z's. One of M's is as warm as 4 of Z's: and one of L's as warm as 3 of M's. Which is best, giving equal weight in the result to rapidity of work, lightness, and warmth?

  Answer.—The order is M, L, Z.

  Solution.—As to rapidity (other things being constant) L's merit is to M's in the ratio of 5 to 2: Z's to L's in the ratio of 4 to 3. In order to get one set of 3 numbers fulfilling these conditions, it is perhaps simplest to take the one that occurs twice as unity, and reduce the others to fractions: this gives, for L, M, and Z, the marks 1, 2⁄5, 2⁄3. In estimating for lightness, we observe that the greater the w
eight, the less the merit, so that Z's merit is to L's as 5 to 1. Thus the marks for lightness are 1⁄5, 2⁄3, 1. And similarly, the marks for warmth are 3, 1, ¼. To get the total result, we must multiply L's 3 marks together, and do the same for M and for Z. The final numbers are 1 × 1⁄5 × 3, 2⁄5 × 2⁄3 × 1, 2⁄3 × 1 × ¼; i.e. 3⁄5, 2⁄3, 1⁄3; i.e. multiplying throughout by 15 (which will not alter the proportion), 9, 10, 5; showing the order of merit to be M, L, Z.

  Twenty-nine answers have been received, of which five are right, and twenty-four wrong. These hapless ones have all (with three exceptions) fallen into the error of adding the proportional numbers together, for each candidate, instead of multiplying. Why the latter is right, rather than the former, is fully proved in text-books, so I will not occupy space by stating it here: but it can be illustrated very easily by the case of length, breadth, and depth. Suppose A and B are rival diggers of rectangular tanks: the amount of work done is evidently measured by the number of cubical feet dug out. Let A dig a tank 10 feet long, 10 wide, 2 deep: let B dig one 6 feet long, 5 wide, 10 deep. The cubical contents are 200, 300; i.e. B is best digger in the ratio of 3 to 2. Now try marking for length, width, and depth, separately; giving a maximum mark of 10 to the best in each contest, and then adding the results!

  Of the twenty-four malefactors, one gives no working, and so has no real claim to be named; but I break the rule for once, in deference to its success in Problem 1: he, she, or it, is Addlepate. The other twenty-three may be divided into five groups.

  First and worst are, I take it, those who put the rightful winner last; arranging them as "Lolo, Zuzu, Mimi." The names of these desperate wrong-doers are Ayr, Bradshaw of the Future, Furze-bush and Pollux (who send a joint answer), Greystead, Guy, Old Hen, and Simple Susan. The latter was once best of all; the Old Hen has taken advantage of her simplicity, and beguiled her with the chaff which was the bane of her own chickenhood.

  Secondly, I point the finger of scorn at those who have put the worst candidate at the top; arranging them as "Zuzu, Mimi, Lolo." They are Graecia, M. M., Old Cat, and R. E. X. "'Tis Greece, but——."

  The third set have avoided both these enormities, and have even succeeded in putting the worst last, their answer being "Lolo, Mimi, Zuzu." Their names are Ayr (who also appears among the "quite too too"), Clifton C., F. B., Fifee, Grig, Janet, and Mrs. Sairey Gamp. F. B. has not fallen into the common error; she multiplies together the proportionate numbers she gets, but in getting them she goes wrong, by reckoning warmth as a de-merit. Possibly she is "Freshly Burnt," or comes "From Bombay." Janet and Mrs. Sairey Gamp have also avoided this error: the method they have adopted is shrouded in mystery—I scarcely feel competent to criticize it. Mrs. Gamp says "if Zuzu makes 4 while Lolo makes 3, Zuzu makes 6 while Lolo makes 5 (bad reasoning), while Mimi makes 2." From this she concludes "therefore Zuzu excels in speed by 1" (i.e. when compared with Lolo; but what about Mimi?). She then compares the 3 kinds of excellence, measured on this mystic scale. Janet takes the statement, that "Lolo makes 5 while Mimi makes 2," to prove that "Lolo makes 3 while Mimi makes 1 and Zuzu 4" (worse reasoning than Mrs. Gamp's), and thence concludes that "Zuzu excels in speed by 1⁄8"! Janet should have been Adeline, "mystery of mysteries!"

  The fourth set actually put Mimi at the top, arranging them as "Mimi, Zuzu, Lolo." They are Marquis and Co., Martreb, S. B. B. (first initial scarcely legible: may be meant for "J"), and Stanza.

  The fifth set consist of An ancient Fish and Camel. These ill-assorted comrades, by dint of foot and fin, have scrambled into the right answer, but, as their method is wrong, of course it counts for nothing. Also An ancient Fish has very ancient and fishlike ideas as to how numbers represent merit: she says "Lolo gains 2½ on Mimi." Two and a half what? Fish, fish, art thou in thy duty?

  Of the five winners I put Balbus and The elder Traveller slightly below the other three—Balbus for defective reasoning, the other for scanty working. Balbus gives two reasons for saying that addition of marks is not the right method, and then adds "it follows that the decision must be made by multiplying the marks together." This is hardly more logical than to say "This is not Spring: therefore it must be Autumn."

  CLASS LIST.

  I.

  Dinah Mite.

  E. B. D. L.

  Joram.

  II.

  Balbus.

  The Elder Traveller.

  With regard to Knot V., I beg to express to Vis Inertiæ and to any others who, like her, understood the condition to be that every marked picture must have three marks, my sincere regret that the unfortunate phrase "fill the columns with oughts and crosses" should have caused them to waste so much time and trouble. I can only repeat that a literal interpretation of "fill" would seem to me to require that every picture in the gallery should be marked. Vis Inertiæ would have been in the First Class if she had sent in the solution she now offers.

  ANSWERS TO KNOT VII.

  Problem.—Given that one glass of lemonade, 3 sandwiches, and 7 biscuits, cost 1s. 2d.; and that one glass of lemonade, 4 sandwiches, and 10 biscuits, cost 1s. 5d.: find the cost of (1) a glass of lemonade, a sandwich, and a biscuit; and (2) 2 glasses of lemonade, 3 sandwiches, and 5 biscuits.

  Answer.—(1) 8d.; (2) 1s. 7d.

  Solution.—This is best treated algebraically. Let x = the cost (in pence) of a glass of lemonade, y of a sandwich, and z of a biscuit. Then we have x + 3y + 7z = 14, and x + 4y + 10z = 17. And we require the values of x + y + z, and of 2x + 3y + 5z. Now, from two equations only, we cannot find, separately, the values of three unknowns: certain combinations of them may, however, be found. Also we know that we can, by the help of the given equations, eliminate 2 of the 3 unknowns from the quantity whose value is required, which will then contain one only. If, then, the required value is ascertainable at all, it can only be by the 3rd unknown vanishing of itself: otherwise the problem is impossible.

  Let us then eliminate lemonade and sandwiches, and reduce everything to biscuits—a state of things even more depressing than "if all the world were apple-pie"—by subtracting the 1st equation from the 2nd, which eliminates lemonade, and gives y + 3z = 3, or y = 3-3z; and then substituting this value of y in the 1st, which gives x-2z = 5, i.e. x = 5 + 2z. Now if we substitute these values of x, y, in the quantities whose values are required, the first becomes (5 + 2z) + (3-3z) + z, i.e. 8: and the second becomes 2(5 + 2z) + 3(3-3z) + 5z, i.e. 19. Hence the answers are (1) 8d., (2) 1s. 7d.

  The above is a universal method: that is, it is absolutely certain either to produce the answer, or to prove that no answer is possible. The question may also be solved by combining the quantities whose values are given, so as to form those whose values are required. This is merely a matter of ingenuity and good luck: and as it may fail, even when the thing is possible, and is of no use in proving it impossible, I cannot rank this method as equal in value with the other. Even when it succeeds, it may prove a very tedious process. Suppose the 26 competitors, who have sent in what I may call accidental solutions, had had a question to deal with where every number contained 8 or 10 digits! I suspect it would have been a case of "silvered is the raven hair" (see "Patience") before any solution would have been hit on by the most ingenious of them.

  Forty-five answers have come in, of which 44 give, I am happy to say, some sort of working, and therefore deserve to be mentioned by name, and to have their virtues, or vices as the case may be, discussed. Thirteen have made assumptions to which they have no right, and so cannot figure in the Class-list, even though, in 10 of the 13 cases, the answer is right. Of the remaining 28, no less than 26 have sent in accidental solutions, and therefore fall short of the highest honours.

  I will now discuss individual cases, taking the worst first, as my custom is.

  Froggy gives no working—at least this is all he gives: after stating the given equations, he says "therefore the difference, 1 sandwich + 3 biscuits, = 3d.": then follow the amounts of the unknown bills, with no further hint as to how he got them. Froggy
has had a very narrow escape of not being named at all!

  Of those who are wrong, Vis Inertiæ has sent in a piece of incorrect working. Peruse the horrid details, and shudder! She takes x (call it "y") as the cost of a sandwich, and concludes (rightly enough) that a biscuit will cost (3-y)/3. She then subtracts the second equation from the first, and deduces 3y + 7 × (3-y)/3-4y + 10 × (3-y)/3 = 3. By making two mistakes in this line, she brings out y = 2⁄2. Try it again, oh Vis Inertiæ! Away with Inertiæ: infuse a little more Vis: and you will bring out the correct (though uninteresting) result, 0 = 0! This will show you that it is hopeless to try to coax any one of these 3 unknowns to reveal its separate value. The other competitor, who is wrong throughout, is either J. M. C. or T. M. C.: but, whether he be a Juvenile Mis-Calculator or a True Mathematician Confused, he makes the answers 7d. and 1s. 5d. He assumes, with Too Much Confidence, that biscuits were ½d. each, and that Clara paid for 8, though she only ate 7!

  We will now consider the 13 whose working is wrong, though the answer is right: and, not to measure their demerits too exactly, I will take them in alphabetical order. Anita finds (rightly) that "1 sandwich and 3 biscuits cost 3d.," and proceeds "therefore 1 sandwich = 1½d., 3 biscuits = 1½d., 1 lemonade = 6d." Dinah Mite begins like Anita: and thence proves (rightly) that a biscuit costs less than a 1d.: whence she concludes (wrongly) that it must cost ½d. F. C. W. is so beautifully resigned to the certainty of a verdict of "guilty," that I have hardly the heart to utter the word, without adding a "recommended to mercy owing to extenuating circumstances." But really, you know, where are the extenuating circumstances? She begins by assuming that lemonade is 4d. a glass, and sandwiches 3d. each, (making with the 2 given equations, four conditions to be fulfilled by three miserable unknowns!). And, having (naturally) developed this into a contradiction, she then tries 5d. and 2d. with a similar result. (N.B. This process might have been carried on through the whole of the Tertiary Period, without gratifying one single Megatherium.) She then, by a "happy thought," tries half-penny biscuits, and so obtains a consistent result. This may be a good solution, viewing the problem as a conundrum: but it is not scientific. Janet identifies sandwiches with biscuits! "One sandwich + 3 biscuits" she makes equal to "4." Four what? Mayfair makes the astounding assertion that the equation, s + 3b = 3, "is evidently only satisfied by s = 2⁄2, b = ½"! Old Cat believes that the assumption that a sandwich costs 1½d. is "the only way to avoid unmanageable fractions." But why avoid them? Is there not a certain glow of triumph in taming such a fraction? "Ladies and gentlemen, the fraction now before you is one that for years defied all efforts of a refining nature: it was, in a word, hopelessly vulgar. Treating it as a circulating decimal (the treadmill of fractions) only made matters worse. As a last resource, I reduced it to its lowest terms, and extracted its square root!" Joking apart, let me thank Old Cat for some very kind words of sympathy, in reference to a correspondent (whose name I am happy to say I have now forgotten) who had found fault with me as a discourteous critic. O. V. L. is beyond my comprehension. He takes the given equations as (1) and (2): thence, by the process [(2)-(1)] deduces (rightly) equation (3) viz. s + 3b = 3: and thence again, by the process [x3] (a hopeless mystery), deduces 3s + 4b = 4. I have nothing to say about it: I give it up. Sea-Breeze says "it is immaterial to the answer" (why?) "in what proportion 3d. is divided between the sandwich and the 3 biscuits": so she assumes s = l½d., b = ½d. Stanza is one of a very irregular metre. At first she (like Janet) identifies sandwiches with biscuits. She then tries two assumptions (s = 1, b = 2⁄3, and s = ½ b = 2⁄6), and (naturally) ends in contradictions. Then she returns to the first assumption, and finds the 3 unknowns separately: quod est absurdum. Stiletto identifies sandwiches and biscuits, as "articles." Is the word ever used by confectioners? I fancied "What is the next article, Ma'am?" was limited to linendrapers. Two Sisters first assume that biscuits are 4 a penny, and then that they are 2 a penny, adding that "the answer will of course be the same in both cases." It is a dreamy remark, making one feel something like Macbeth grasping at the spectral dagger. "Is this a statement that I see before me?" If you were to say "we both walked the same way this morning," and I were to say "one of you walked the same way, but the other didn't," which of the three would be the most hopelessly confused? Turtle Pyate (what is a Turtle Pyate, please?) and Old Crow, who send a joint answer, and Y. Y., adopt the same method. Y. Y. gets the equation s + 3b = 3: and then says "this sum must be apportioned in one of the three following ways." It may be, I grant you: but Y. Y. do you say "must"? I fear it is possible for Y. Y. to be two Y's. The other two conspirators are less positive: they say it "can" be so divided: but they add "either of the three prices being right"! This is bad grammar and bad arithmetic at once, oh mysterious birds!