of which Legendre gave at some length a "proof" of impossibility; but I have since found that Lucas anticipated me in a communication to Sylvester.
The illustration shows how the sixteen trees might have been planted so as to form as many as fifteen straight rows with four trees in every row. This is in excess of what was for a long timebelieved to be the maximum number of rows possible; and though with our present knowledge I cannot rigorously demonstrate that fifteen rows cannot be beaten, I have a strong "pious opinion" that it is the highest number of rows obtainable.
The answer to this puzzle is shown in the illustration, where the numbers on the sixteen bottles all add up to 30 in the ten straight directions. The trick consists in the fact that, although the six bottles (3, 5, 6, 9, 10, and 15) in which the flowers have been placed are not removed, yet the sixteen need not occupy exactly the same position on the table as before. The square is, in fact, formed one step further to the left.
The portrait may be drawn in a single line because it contains only two points at which an odd number of lines meet, but it is absolutely necessary to begin at one of these points and end at the other. One point is near the outer extremity of the King's left eye; the other is below it on the left cheek.
The five hundred silver pennies might have been placed in the four bags, in accordance with the stated conditions, in exactly 894,348 different ways. If there had been a thousand coins there would be 7,049,112 ways. It is a difficult problem in the partition of numbers. I have a single formula for the solution of any number of coins in the case of four bags, but it was extremely hard to construct, and the best method is to find the twelve separate formulas for the different congruences to the modulus 12.
A very little examination of the original drawing will have shown the reader that, as he will have at first read the conditions, the puzzle is quite impossible of solution. We have therefore tolook for some loophole in the actual conditions as they were worded. If the Parson could get round the source of the river, he could then cross every bridge once and once only on his way to church, as shown in the annexed illustration. That this was not prohibited we shall soon find. Though the plan showed all the bridges in his parish, it only showed "part of" the parish itself. It is not stated that the river did not take its rise in the parish, and since it leads to the only possible solution, we must assume that it did. The answer would be, therefore, as shown. It should be noted that we are clearly prevented from considering the possibility of getting round the mouth of the river, because we are told it "joined the sea some hundred miles to the south," while no parish ever extended a hundred miles!
The illustration will show how the triangular piece of cloth may be cut into four pieces that will fit together and form a perfect square. Bisect AB in D and BC in E; produce the line AE to F making EF equal to EB; bisect AF in G and describe thearc AHF; produce EB to H, and EH is the length of the side of the required square; from E with distance EH, describe the arc HJ, and make JK equal to BE; now, from the points D and K drop perpendiculars on EJ at L and M. If you have done this accurately, you will now have the required directions for the cuts.
I exhibited this problem before the Royal Society, at Burlington House, on 17th May 1905, and also at the Royal Institution in the following month, in the more general form:—"A New Problem on Superposition: a demonstration that an equilateral triangle can be cut into four pieces that may be reassembled to form a square, with some examples of a general method for transforming all rectilinear triangles into squares by dissection." It was also issued as a challenge to the readers of theDaily Mail(see issues of 1st and 8th February 1905), but though many hundreds of attempts were sent in there was not a single solver. Credit, however, is due to Mr. C. W. M'Elroy, who alone sent me the correct solution when I first published the problem in theWeekly Dispatchin 1902.
I add an illustration showing the puzzle in a rather curiouspractical form, as it was made in polished mahogany with brass hinges for use by certain audiences. It will be seen that the four pieces form a sort of chain, and that when they are closed up in one direction they form the triangle, and when closed in the other direction they form the square.
The correct answer is 18,816 different ways. The general formula for six fleurs-de-lys for all squares greater than 22is simply this: Six times the square of the number of combinations ofnthings, taken three at a time, wherenrepresents the number of fleurs-de-lys in the side of the square. Of course wherenis even the remainders in rows and columns will be even, and wherenis odd the remainders will be odd.
For further solution, see No. 358 inA. in M.
In this little problem we attempted to show how, by sophistical reasoning, it may apparently be proved that the diagonal of a square is of precisely the same length as two of the sides. The puzzle was to discover the fallacy, because it is a very obvious fallacy if we admit that the shortest distance between two points is a straight line. But where does the error come in?
Well, it is perfectly true that so long as our zigzag path is formed of "steps" parallel to the sides of the square that path must be of the same length as the two sides. It does not matter if you have to use the most powerful microscope obtainable; the rule is always true if the path is made up of steps in that way. But the error lies in the assumption that such a zigzag path can ever become a straight line. You may go on increasing the number of steps infinitely—that is, there is no limit whatever theoretically to the number of steps that can be made—but you can never reach a straight line by such a method. In fact it is just as much a "jump" to a straight line if you have a billion steps as it is atthe very outset to pass from the two sides to the diagonal. It would be just as absurd to say we might go on dropping marbles into a basket until they become sovereigns as to say we can increase the number of our steps until they become a straight line. There is the whole thing in a nutshell.
The surface of water or other liquid is always spherical, and the greater any sphere is the less is its convexity. Hence the top diameter of any vessel at the summit of a mountain will form the base of the segment of a greater sphere than it would at the bottom. This sphere, being greater, must (from what has been already said) be less convex; or, in other words, the spherical surface of the water must be less above the brim of the vessel, and consequently it will hold less at the top of a mountain than at the bottom. The reader is therefore free to select any mountain he likes in Italy—or elsewhere!
The number of different ways is 63,504. The general formula for such arrangements, when the number of letters in the sentence is 2n+ 1, and it is a palindrome without diagonal readings, is [4(2n- 1)]2.
I think it will be well to give here a formula for the general solution of each of the four most common forms of the diamond-letter puzzle. By the word "line" I mean the complete diagonal. Thus in A, B, C, and D, the lines respectively contain 5, 5, 7, and 9 letters. A has a non-palindrome line (the word being BOY), and the general solution for such cases, where the line contains 2n+ 1 letters, is 4(2n- 1). Where the line is a single palindrome, with its middle letter in the centre, as in B, the general formula is [4(2n- 1)]2. This is the form of the Rat-catcher's Puzzle, and therefore the expression that I have given above. In cases C and D we have double palindromes, but these two represent verydifferent types. In C, where the line contains 4n-1 letters, the general expression is 4(22n-2). But D is by far the most difficult case of all.
I had better here state that in the diamonds under consideration (i.) no diagonal readings are allowed—these have to be dealt with specially in cases where they are possible and admitted; (ii.) readings may start anywhere; (iii.) readings may go backwards and forwards, using letters more than once in a single reading, but not the same letter twice in immediate succession. This last condition will be understood if the reader glances at C, where it is impossible to go forwards and backwards in a reading without repeating the first O touched—a proceeding which I have said is not allowed. In the case D it is very different, and this is what accounts for its greater difficulty. The formula for D is this:
where the number of letters in the line is 4n+ 1. In the example given there are therefore 400 readings forn= 2.
See also Nos. 256, 257, and 258 inA. in M.
The simple Ploughman, who was so ridiculed for his opinion, was perfectly correct: the Miller should receive seven pieces of money, and the Weaver only one. As all three ate equal shares of the bread, it should be evident that each ate 8/3 of a loaf. Therefore, as the Miller provided 15/3 and ate 8/3, he contributed 7/3 to the Manciple's meal; whereas the Weaver provided 9/3, ate 8/3, and contributed only 1/3. Therefore, since they contributed to the Manciple in the proportion of 7 to 1, they must divide the eight pieces of money in the same proportion.
SIR HUGH EXPLAINS HIS PROBLEMS
The friends of Sir Hugh de Fortibus were so perplexed over many of his strange puzzles that at a gathering of his kinsmen and retainers he undertook to explain his posers.
"Of a truth," said he, "some of the riddles that I have put forth would greatly tax the wit of the unlettered knave to rede; yet will I try to show the manner thereof in such way that all may have understanding. For many there be who cannot of themselvesdo all these things, but will yet study them to their gain when they be given the answers, and will take pleasure therein."
Sir Hugh explained, in answer to this puzzle, that as the nine holes were 300, 250, 200, 325, 275, 350, 225, 375, and 400 yards apart, if a man could always strike the ball in a perfectly straight line and send it at will a distance of either 125 yards or 100 yards, he might go round the whole course in 26 strokes. This is clearly correct, for if we call the 125 stroke the "drive" and the 100 stroke the "approach," he could play as follows:—The first hole could be reached in 3 approaches, the second in 2 drives, the third in 2 approaches, the fourth in 2 approaches and 1 drive, the fifth in 3 drives and 1 backward approach, the sixth in 2 drives and 1 approach, the seventh in 1 drive and 1 approach, the eighth in 3 drives, and the ninth hole in 4 approaches. There are thus 26 strokes in all, and the feat cannot be performed in fewer.
"By my halidame!" exclaimed Sir Hugh, "if some of yon varlets had been put in chains, which for their sins they do truly deserve, then would they well know, mayhap, that the length of any chain having like rings is equal to the inner width of a ring multiplied by the number of rings and added to twice the thickness of the iron whereof it is made. It may be shown that the inner width of the rings used in the tilting was one inch and two-thirdsthereof, and the number of rings Stephen Malet did win was three, and those that fell to Henry de Gournay would be nine."
The knight was quite correct, for 1-2/3 in. × 3 + 1 in. = 6 in., and 1-2/3 in. x 9 + 1 in. = 16 in. Thus De Gournay beat Malet by six rings. The drawing showing the rings may assist the reader in verifying the answer and help him to see why the inner width of a link multiplied by the number of links and added to twice the thickness of the iron gives the exact length. It will be noticed that every link put on the chain loses a length equal to twice the thickness of the iron.
"Some here have asked me," continued Sir Hugh, "how they may find the cell in the Dungeon of the Death's-head wherein the noble maiden was cast. Beshrew me! but 'tis easy withal when you do but know how to do it. In attempting to pass through every door once, and never more, you must take heed that every cell hath two doors or four, which be even numbers, except two cells, which have but three. Now, certes, you cannot go in and out of any place, passing through all the doors once and no more, if the number of doors be an odd number. But as there be but two such odd cells, yet may we, by beginning at the one and ending at the other, so make our journey in many ways with success. I pray you, albeit, to mark that only one of these odd cells lieth onthe outside of the dungeon, so we must perforce start therefrom. Marry, then, my masters, the noble demoiselle must needs have been wasting in the other."
The drawing will make this quite clear to the reader. The two "odd cells" are indicated by the stars, and one of the many routes that will solve the puzzle is shown by the dotted line. It is perfectly certain that you must start at the lower star and end at the upper one; therefore the cell with the star situated over the left eye must be the one sought.
"It hath been said that the proof of a pudding is ever in the eating thereof, and by the teeth of Saint George I know no better way of showing how this placing of the figures may be done than by the doing of it. Therefore have I in suchwise written the numbersthat they do add up to twenty and three in all the twelve lines of three that are upon the butt."
I think it well here to supplement the solution of De Fortibus with a few remarks of my own. The nineteen numbers may be so arranged that the lines will add up to any number we may choose to select from 22 to 38 inclusive, excepting 30. In some cases there are several different solutions, but in the case of 23 there are only two. I give one of these. To obtain the second solution exchange respectively 7, 10, 5, 8, 9, in the illustration, with 13, 4, 17, 2, 15. Also exchange 18 with 12, and the other numbers may remain unmoved. In every instance there must be an even number in the central place, and any such number from 2 to 18 may occur. Every solution has its complementary. Thus, if for every number in the accompanying drawing we substitute the difference between it and 20, we get the solution in the case of 37. Similarly, from the arrangement in the original drawing, we may at once obtain a solution for the case of 38.
In this case Sir Hugh had greatly perplexed his chief builder by demanding that he should make a window measuring one foot on every side and divided by bars into eight lights, having all their sides equal. The illustration will show how this was to bedone. It will be seen that if each side of the window measures one foot, then each of the eight triangular lights is six inches on every side.
"Of a truth, master builder," said De Fortibus slyly to the architect, "I did not tell thee that the window must be square, as it is most certain it never could be."
"By the toes of St. Moden," exclaimed Sir Hugh de Fortibus when this puzzle was brought up, "my poor wit hath never shaped a more cunning artifice or any more bewitching to look upon. It came to me as in a vision, and ofttimes have I marvelled at the thing, seeing its exceeding difficulty. My masters and kinsmen, it is done in this wise."
The worthy knight then pointed out that the crescent was of a particular and somewhat irregular form—the two distancesatobandctodbeing straight lines, and the arcsacandbdbeing precisely similar. He showed that if the cuts be made as in Figure 1, the four pieces will fit together and form a perfect square, as shown in Figure 2, if we there only regard the three curved lines. By now making the straight cuts also shown in Figure 2, we get the ten pieces that fit together, as in Figure 3, and form a perfectly symmetrical Greek cross. The proportions of the crescent andthe cross in the original illustration were correct, and the solution can be demonstrated to be absolutely exact and not merely approximate.
I have a solution in considerably fewer pieces, but it is far more difficult to understand than the above method, in which the problem is simplified by introducing the intermediate square.
The puzzle was to place your pencil on the A at the top of the amulet and count in how many different ways you could trace out the word "Abracadabra" downwards, always passing from a letter to an adjoining one.
"Now, mark ye, fine fellows," said Sir Hugh to some who had besought him to explain, "that at the very first start there be two ways open: whichever B ye select, there will be two several ways of proceeding (twice times two are four); whichever R ye select, there be two ways of going on (twice times four are eight); and so on until the end. Each letter in order from A downwards may so be reached in 2, 4, 8, 16, 32, etc., ways. Therefore, as there be ten lines or steps in all from A to the bottom, all ye need do is to multiply ten 2's together, and truly the result, 1024, is the answer thou dost seek."
Though there was no need to take down and measure the staff, it is undoubtedly necessary to find its height before the answercan be given. It was well known among the friends and retainers of Sir Hugh de Fortibus that he was exactly six feet in height. It will be seen in the original picture that Sir Hugh's height is just twice the length of his shadow. Therefore we all know that the flagstaff will, at the same place and time of day, be also just twice as long as its shadow. The shadow of the staff is the same length as Sir Hugh's height; therefore this shadow is six feet long, and the flagstaff must be twelve feet high. Now, the snail, by climbing up three feet in the daytime and slipping back two feet by night, really advances one foot in a day of twenty-four hours. At the end of nine days it is three feet from the top, so that it reaches its journey's end on the tenth day.
The reader will doubtless here exclaim, "This is all very well; but how were we to know the height of Sir Hugh? It was never stated how tall he was!" No, it was not stated in so many words, but it was none the less clearly indicated to the reader who is sharp in these matters. In the original illustration to the donjon keep window Sir Hugh is shown standing against a wall, the window in which is stated to be one foot square on the inside. Therefore, as his height will be found by measurement to be just six times the inside height of the window, he evidently stands just six feet in his boots!
The last puzzle was undoubtedly a hard nut, but perhaps difficulty does not make a good puzzle any the less interesting when we are shown the solution. The accompanying diagram indicates exactly how the top of Lady Isabel de Fitzarnulph's casket was inlaid with square pieces of rare wood (no two squares alike) and the strip of gold 10 inches by a quarter of an inch. This is the only possible solution, and it is a singular fact (though I cannot here show the subtle method of working) that the number, sizes, and order of those squares are determined by the given dimensions of the strip of gold, and the casket can have no other dimensions than 20 inches square. The number in a square indicates the lengthin inches of the side of that square, so the accuracy of the answer can be checked almost at a glance.
Sir Hugh de Fortibus made some general concluding remarks on the occasion that are not altogether uninteresting to-day.
"Friends and retainers," he said, "if the strange offspring of my poor wit about which we have held pleasant counsel to-night hath mayhap had some small interest for ye, let these matters serve to call to mind the lesson that our fleeting life is rounded and beset with enigmas. Whence we came and whither we go be riddles, and albeit such as these we may never bring within our understanding, yet there be many others with which we and they thatdo come after us will ever strive for the answer. Whether success do attend or do not attend our labour, it is well that we make the attempt; for 'tis truly good and honourable to train the mind, and the wit, and the fancy of man, for out of such doth issue all manner of good in ways unforeseen for them that do come after us."
Number the fish baskets in the illustration from 1 to 12 in the direction that Brother Jonathan is seen to be going. Starting from 1, proceed as follows, where "1 to 4" means, take the fish from basket No. 1 and transfer it to basket No. 4:—
1 to 4, 5 to 8, 9 to 12, 3 to 6, 7 to 10, 11 to 2, and complete the last revolution to 1, making three revolutions in all. Or you can proceed this way:—
4 to 7, 8 to 11, 12 to 3, 2 to 5, 6 to 9, 10 to 1.
It is easy to solve in four revolutions, but the solutions in three are more difficult to discover.
If it were not for the Abbot's conditions that the number of guests in any room may not exceed three, and that every room must be occupied, it would have been possible to accommodate either 24, 27, 30, 33, 36, 39, or 42 pilgrims. But to accommodate 24 pilgrims so that there shall be twice as many sleeping on the upper floor as on the lower floor, and eleven persons on each side of the building, it will be found necessary to leave some of the rooms empty. If, on the other hand, we try to put up 33, 36, 39 or 42 pilgrims, we shall find that in every case we are obliged to place more than three persons in some of the rooms. Thus we know that the number of pilgrims originally announced (whom, it will be remembered, it was possible to accommodate under theconditions of the Abbot) must have been 27, and that, since three more than this number were actually provided with beds, the total number of pilgrims was 30. The accompanying diagram shows how they might be arranged, and if in each instance we regard the upper floor as placed above the lower one, it will be seen that there are eleven persons on each side of the building, and twice as many above as below.
The correct answer is shown in the illustration on page 196. No tile is in line (either horizontally, vertically, or diagonally) with another tile of the same design, and only three plain tiles are used. If after placing the four lions you fall into the error of placing four other tiles of another pattern, instead of only three, you will be left with four places that must be occupied by plain tiles. The secret consists in placing four of one kind and only three of each of the others.
The question was: Did Brother Benjamin take more wine from the bottle than water from the jug? Or did he take more water from the jug than wine from the bottle? He did neither. The same quantity of wine was transferred from the bottle as water was taken from the jug. Let us assume that the glass would hold a quarter of a pint. There was a pint of wine in the bottle and a pint of water in the jug. After the first manipulation the bottle contains three-quarters of a pint of wine, and the jug one pint of water mixed with a quarter of a pint of wine. Now, the second transaction consists in taking away a fifth of the contents of the jug—that is, one-fifth of a pint of water mixed with one-fifth of a quarter of a pint of wine. We thus leave behind in the jug four-fifths of a quarter of a pint of wine—that is, one-fifth of a pint—while we transfer from the jug to the bottle an equal quantity (one-fifth of a pint) of water.
There were 100 pints of wine in the cask, and on thirty occasions John the Cellarer had stolen a pint and replaced it with a pint of water. After the first theft the wine left in the cask would be99 pints; after the second theft the wine in the cask would be 9801/100 pints (the square of 99 divided by 100); after the third theft there would remain 970299/10000 (the cube of 99 divided by the square of 100); after the fourth theft there would remain the fourth power of 99 divided by the cube of 100; and after the thirtieth theft there would remain in the cask the thirtieth power of 99 divided by the twenty-ninth power of 100. This by the ordinary method of calculation gives us a number composed of 59 figures to be divided by a number composed of 58 figures! But by the use of logarithms it may be quickly ascertained that the required quantity is very nearly 73-97/100 pints of wine left in the cask. Consequently the cellarer stole nearly 26.03 pints. The monks doubtless omitted the answer for the reason that they had no tables of logarithms, and did not care to face the task of making that long and tedious calculation in order to get the quantity "to a nicety," as the wily cellarer had stipulated.
By a simplified process of calculation, I have ascertained that the exact quantity of wine stolen would be
26.0299626611719577269984907683285057747323737647323555652999
pints. A man who would involve the monastery in a fraction of fifty-eight decimals deserved severe punishment.
The correct answer is that there would have been 602,176 Crusaders, who could form themselves into a square 776 by 776; and after the stranger joined their ranks, they could form 113 squares of 5,329 men—that is, 73 by 73. Or 113 × 732- 1 = 7762. This is a particular case of the so-called "Pellian Equation," respecting which seeA. in M., p. 164.
The reader is aware that there are prime numbers and composite whole numbers. Now, 1,111,111 cannot be a prime number,because if it were the only possible answers would be those proposed by Brother Benjamin and rejected by Father Peter. Also it cannot have more than two factors, or the answer would be indeterminate. As a matter of fact, 1,111,111 equals 239 x 4649 (both primes), and since each cat killed more mice than there were cats, the answer must be 239 cats. See also the Introduction, p.18.
Treated generally, this problem consists in finding the factors, if any, of numbers of the form (10n- 1)/9.
Lucas, in hisL'Arithmétique Amusante, gives a number of curious tables which he obtained from an arithmetical treatise, called theTalkhys, by Ibn Albanna, an Arabian mathematician and astronomer of the first half of the thirteenth century. In the Paris National Library are several manuscripts dealing with theTalkhys, and a commentary by Alkalaçadi, who died in 1486. Among the tables given by Lucas is one giving all the factors of numbers of the above form up ton= 18. It seems almost inconceivable that Arabians of that date could find the factors wheren= 17, as given in my Introduction. But I read Lucas as stating that they are given inTalkhys, though an eminent mathematician reads him differently, and suggests to me that they were discovered by Lucas himself. This can, of course, be settled by an examination ofTalkhys, but this has not been possible during the war.
The difficulty lies wholly with those cases wherenis a prime number. Ifn= 2, we get the prime 11. The factors whenn= 3, 5, 11, and 13 are respectively (3 . 37), (41 . 271), (21,649 . 513,239), and (53 . 79 . 265371653). I have given in these pages the factors wheren= 7 and 17. The factors whenn= 19, 23, and 37 are unknown, if there are any.[B]Whenn= 29, the factors are (3,191 . 16,763 . 43,037.62,003 . 77,843,839,397); whenn= 31, one factor is 2,791; and whenn= 41, two factors are (83 . 1,231).
[B]Mr. Oscar Hoppe, of New York, informs me that, after reading my statement in the Introduction, he was led to investigate the case ofn= 19, and after long and tedious work he succeeded in proving the number to be a prime. He submitted his proof to the London Mathematical Society, and a specially appointed committee of that body accepted the proof as final and conclusive. He refers me to theProceedingsof the Society for 14th February 1918.
[B]Mr. Oscar Hoppe, of New York, informs me that, after reading my statement in the Introduction, he was led to investigate the case ofn= 19, and after long and tedious work he succeeded in proving the number to be a prime. He submitted his proof to the London Mathematical Society, and a specially appointed committee of that body accepted the proof as final and conclusive. He refers me to theProceedingsof the Society for 14th February 1918.
As for the even values ofn, the following curious series of factors will doubtless interest the reader. The numbers in brackets are primes.
n= 2 = (11)
n= 6 = (11) × 111 × 91
n= 10 = (11) × 11,111 × (9,091)
n= 14 = (11) × 1,111,111 × (909,091)
n= 18 = (11) × 111,111,111 × 90,909,091
Or we may put the factors this way:—
n= 2 = (11)
n= 6 = 111 × 1,001
n= 10 = 11,111 × 100,001
n= 14 = 1,111,111 × 10,000,001
n= 18 = 111,111,111 × 1,000,000,001
In the above two tablesnis of the form 4m+ 2. Whennis of the form 4mthe factors may be written down as follows:—
n= 4 = (11) × (101)
n= 8 = (11) × (101) × 10,001
n= 12 = (11) × (101) × 100,010,001
n= 16 = (11) × (101) × 1,000,100,010,001.
Whenn= 2, we have the prime number 11; whenn= 3, the factors are 3 . 37; whenn= 6, they are 11 . 3 . 37 . 7. 13; whenn= 9, they are 32. 37 . 333,667. Therefore we know that factors ofn= 18 are 11. 32. 37 . 7 . 13 . 333,667, while the remaining factor is composite and can be split into 19 . 52579. This will show how the working may be simplified whennis not prime.
The fewest possible moves in which this puzzle can be solved are 118. I will give the complete solution. The black figures onwhite discs move in the directions of the hands of a clock, and the white figures on black discs the other way. The following are the numbers in the order in which they move. Whether you have to make a simple move or a leaping move will be clear from the position, as you never can have an alternative. The moves enclosed in brackets are to be played five times over: 6, 7, 8, 6, 5, 4, 7, 8, 9, 10, 6, 5, 4, 3, 2, 7, 8, 9, 10, 11 (6, 5, 4, 3, 2, 1), 6, 5, 4, 3, 2, 12, (7, 8, 9, 10, 11, 12), 7, 8, 9, 10, 11, 1, 6, 5, 4, 3, 2, 12, 7, 8, 9, 10, 11, 6, 5, 4, 3, 2, 8, 9, 10, 11, 4, 3, 2, 10, 11, 2. We thus have made 118 moves within the conditions, the black frogs have changed places with the white ones, and 1 and 12 are side by side in the positions stipulated.
The general solution in the case of this puzzle is 3n2+ 2n- 2 moves, where the number of frogs of each colour isn. The law governing the sequence of moves is easily discovered by an examination of the simpler cases, wheren= 2, 3, and 4.
If, instead of 11 and 12 changing places, the 6 and 7 must interchange, the expression isn2+ 4n+ 2 moves. If we giventhe value 6, as in the example of the Frogs' Ring, the number of moves would be 62.
For a general solution of the case where frogs of one colour reverse their order, leaving the blank space in the same position, and each frog is allowed to be moved in either direction (leaping, of course, over his own colour), see "The Grasshopper Puzzle" inA. in M., p. 193.
Although the king's jester promised that he would "thereafter make the manner thereof plain to all," there is no record of his having ever done so. I will therefore submit to the reader my own views as to the probable solutions to the mysteries involved.
When the jester "divided his rope in half," it does not follow that he cut it into two parts, each half the original length of the rope. No doubt he simply untwisted the strands, and so divided it into two ropes, each of the original length, but one-half the thickness. He would thus be able to tie the two together and make a rope nearly twice the original length, with which it is quite conceivable that he made good his escape from the dungeon.
How did the jester find his way out of the maze in the dark? He had simply to grope his way to a wall and then keep on walking without once removing his left hand (or right hand) from the wall. Starting from A, the dotted line will make the route clear when he goes to the left. If the reader tries the route to the right in the same way he will be equally successful; in fact, the two routes unite and cover every part of the walls of the maze except those two detached parts on the left-hand side—one piece like aU, and the other like a distorted E. This rule will apply to the majority of mazes and puzzle gardens; but if the centre were enclosed by an isolated wall in the form of a split ring, the jester would simply have gone round and round this ring.