Amusements in Mathematics Part 29

Now, there are just eight different first half-strings, and consequently also eight second half-strings. We shall see that these combine to form twelve complete strings, which is the total number that exist and the correct solution of our puzzle. I do not propose to give all the routes at length, but I will so far indicate them that if the reader has dropped any he will be able to discover which they are and work them out for himself without any difficulty. The following numbers apply to those in the above diagram.

The eight first half-strings are: 1 to 6 (2 routes); 1 to 8 (1 route); 1 to 10 (3 routes); 1 to 12 (1 route); and 1 to 14 (1 route). The eight second half-strings are: 7 to 20 (1 route); 9 to 20 (1 route); 11 to 20 (3 routes); 13 to 20 (1 route); and 15 to 20 (2 routes). Every different way in which you can link one half-string to another gives a different solution. These linkings will be found to be as follows: 6 to 13 (2 cases); 10 to 13 (3 cases); 8 to 11 (3 cases); 8 to 15 (2 cases); 12 to 9 (1 case); and 14 to 7 (1 case). There are, therefore, twelve different linkings and twelve different answers to the puzzle. The route given in the ill.u.s.tration with the greyhound will be found to consist of one of the three half-strings 1 to 10, linked to the half-string 13 to 20. It should be noted that ten of the solutions are produced by five distinctive routes and their reversals--that is, if you indicate these five routes by lines and then turn the diagrams upside down you will get the five other routes. The remaining two solutions are symmetrical (these are the cases where 12 to 9 and 14 to 7 are the links), and consequently they do not produce new solutions by reversal.

337.--THE FOUR KANGAROOS.

[Ill.u.s.tration]

A pretty symmetrical solution to this puzzle is shown in the diagram. Each of the four kangaroos makes his little excursion and returns to his corner, without ever entering a square that has been visited by another kangaroo and without crossing the central line. It will at once occur to the reader, as a possible improvement of the puzzle, to divide the board by a central vertical line and make the condition that this also shall not be crossed. This would mean that each kangaroo had to confine himself to a square 4 by 4, but it would be quite impossible, as I shall explain in the next two puzzles.

338.--THE BOARD IN COMPARTMENTS.

[Ill.u.s.tration]

In attempting to solve this problem it is first necessary to take the two distinctive compartments of twenty and twelve squares respectively and a.n.a.lyse them with a view to determining where the necessary points of entry and exit lie. In the case of the larger compartment it will be found that to complete a tour of it we must begin and end on two of the outside squares on the long sides. But though you may start at any one of these ten squares, you are restricted as to those at which you can end, or (which is the same thing) you may end at whichever of these you like, provided you begin your tour at certain particular squares. In the case of the smaller compartment you are compelled to begin and end at one of the six squares lying at the two narrow ends of the compartments, but similar restrictions apply as in the other instance. A very little thought will show that in the case of the two small compartments you must begin and finish at the ends that lie together, and it then follows that the tours in the larger compartments must also start and end on the contiguous sides.

In the diagram given of one of the possible solutions it will be seen that there are eight places at which we may start this particular tour; but there is only one route in each case, because we must complete the compartment in which we find ourself before pa.s.sing into another. In any solution we shall find that the squares distinguished by stars must be entering or exit points, but the law of reversals leaves us the option of making the other connections either at the diamonds or at the circles. In the solution worked out the diamonds are used, but other variations occur in which the circle squares are employed instead. I think these remarks explain all the essential points in the puzzle, which is distinctly instructive and interesting.

339.--THE FOUR KNIGHTS' TOURS.

[Ill.u.s.tration]

It will be seen in the ill.u.s.tration how a chessboard may be divided into four parts, each of the same size and shape, so that a complete re-entrant knight's tour may be made on each portion. There is only one possible route for each knight and its reversal.

340.--THE CUBIC KNIGHT'S TOUR.

[Ill.u.s.tration]

If the reader should cut out the above diagram, fold it in the form of a cube, and stick it together by the strips left for that purpose at the edges, he would have an interesting little curiosity. Or he can make one on a larger scale for himself. It will be found that if we imagine the cube to have a complete chessboard on each of its sides, we may start with the knight on any one of the 384 squares, and make a complete tour of the cube, always returning to the starting-point. The method of pa.s.sing from one side of the cube to another is easily understood, but, of course, the difficulty consisted in finding the proper points of entry and exit on each board, the order in which the different boards should be taken, and in getting arrangements that would comply with the required conditions.

341.--THE FOUR FROGS.

The fewest possible moves, counting every move separately, are sixteen. But the puzzle may be solved in seven plays, as follows, if any number of successive moves by one frog count as a single play. All the moves contained within a bracket are a single play; the numbers refer to the toadstools: (1--5), (3--7, 7--1), (8--4, 4--3, 3--7), (6--2, 2--8, 8--4, 4--3), (5--6, 6--2, 2--8), (1--5, 5--6), (7--1).

This is the familiar old puzzle by Guarini, propounded in 1512, and I give it here in order to explain my "b.u.t.tons and string" method of solving this cla.s.s of moving-counter problem.

Diagram A shows the old way of presenting Guarini's puzzle, the point being to make the white knights change places with the black ones. In "The Four Frogs" presentation of the idea the possible directions of the moves are indicated by lines, to obviate the necessity of the reader's understanding the nature of the knight's move in chess. But it will at once be seen that the two problems are identical. The central square can, of course, be ignored, since no knight can ever enter it. Now, regard the toadstools as b.u.t.tons and the connecting lines as strings, as in Diagram B. Then by disentangling these strings we can clearly present the diagram in the form shown in Diagram C, where the relations.h.i.+p between the b.u.t.tons is precisely the same as in B. Any solution on C will be applicable to B, and to A. Place your white knights on 1 and 3 and your black knights on 6 and 8 in the C diagram, and the simplicity of the solution will be very evident. You have simply to move the knights round the circle in one direction or the other. Play over the moves given above, and you will find that every little difficulty has disappeared.

[Ill.u.s.trations: A B C D E]

In Diagram D I give another familiar puzzle that first appeared in a book published in Brussels in 1789, Les Pet.i.tes Aventures de Jerome Sharp. Place seven counters on seven of the eight points in the following manner. You must always touch a point that is vacant with a counter, and then move it along a straight line leading from that point to the next vacant point (in either direction), where you deposit the counter. You proceed in the same way until all the counters are placed. Remember you always touch a vacant place and slide the counter from it to the next place, which must be also vacant. Now, by the "b.u.t.tons and string" method of simplification we can transform the diagram into E. Then the solution becomes obvious. "Always move to the point that you last moved from." This is not, of course, the only way of placing the counters, but it is the simplest solution to carry in the mind.

There are several puzzles in this book that the reader will find lend themselves readily to this method.

342.--THE MANDARIN'S PUZZLE.

The rather perplexing point that the solver has to decide for himself in attacking this puzzle is whether the shaded numbers (those that are shown in their right places) are mere dummies or not. Ninety-nine persons out of a hundred might form the opinion that there can be no advantage in moving any of them, but if so they would be wrong.

The shortest solution without moving any shaded number is in thirty-two moves. But the puzzle can be solved in thirty moves. The trick lies in moving the 6, or the 15, on the second move and replacing it on the nineteenth move. Here is the solution: 2, 6, 13, 4, 1, 21, 4, 1, 10, 2, 21, 10, 2, 5, 22, 16, 1, 13, 6, 19, 11, 2, 5, 22, 16, 5, 13, 4, 10, 21. Thirty moves.

343.--EXERCISE FOR PRISONERS.

There are eighty different arrangements of the numbers in the form of a perfect knight's path, but only forty of these can be reached without two men ever being in a cell at the same time. Two is the greatest number of men that can be given a complete rest, and though the knight's path can be arranged so as to leave either 7 and 13, 8 and 13, 5 and 7, or 5 and 13 in their original positions, the following four arrangements, in which 7 and 13 are unmoved, are the only ones that can be reached under the moving conditions. It therefore resolves itself into finding the fewest possible moves that will lead up to one of these positions. This is certainly no easy matter, and no rigid rules can be laid down for arriving at the correct answer. It is largely a matter for individual judgment, patient experiment, and a sharp eye for revolutions and position.

A +--+--+--+--+ | 6| 1|10|15| +--+--+--+--+ | 9|12| 7| 4| +--+--+--+--+ | 2| 5|14|11| +--+--+--+--+ |13| 8| 3|**| +--+--+--+--+ B +--+--+--+--+ | 6| 1|10|15| +--+--+--+--+ |11|14| 7| 4| +--+--+--+--+ | 2| 5|12| 9| +--+--+--+--+ |13| 8| 3|**| +--+--+--+--+ C +--+--+--+--+ | 6| 9| 4|15| +--+--+--+--+ | 1|12| 7|10| +--+--+--+--+ | 8| 5|14| 3| +--+--+--+--+ |13| 2|11|**| +--+--+--+--+ D +--+--+--+--+ | 6|11| 4|15| +--+--+--+--+ | 1|14| 7|10| +--+--+--+--+ | 8| 5|12| 3| +--+--+--+--+ |13| 2| 9|**| +--+--+--+--+ [Ill.u.s.tration: A, B, C, D]

As a matter of fact, the position C can be reached in as few as sixty-six moves in the following manner: 12, 11, 15, 12, 11, 8, 4, 3, 2, 6, 5, 1, 6, 5, 10, 15, 8, 4, 3, 2, 5, 10, 15, 8, 4, 3, 2, 5, 10, 15, 8, 4, 12, 11, 3, 2, 5, 10, 15, 6, 1, 8, 4, 9, 8, 1, 6, 4, 9, 12, 2, 5, 10, 15, 4, 9, 12, 2, 5, 3, 11, 14, 2, 5, 14, 11 = 66 moves. Though this is the shortest that I know of, and I do not think it can be beaten, I cannot state positively that there is not a shorter way yet to be discovered. The most tempting arrangement is certainly A; but things are not what they seem, and C is really the easiest to reach.

If the bottom left-hand corner cell might be left vacant, the following is a solution in forty-five moves by Mr. R. Elrick: 15, 11, 10, 9, 13, 14, 11, 10, 7, 8, 4, 3, 8, 6, 9, 7, 12, 4, 6, 9, 5, 13, 7, 5, 13, 1, 2, 13, 5, 7, 1, 2, 13, 8, 3, 6, 9, 12, 7, 11, 14, 1, 11, 14, 1. But every man has moved.

344.--THE KENNEL PUZZLE.

The first point is to make a choice of the most promising knight's string and then consider the question of reaching the arrangement in the fewest moves. I am strongly of opinion that the best string is the one represented in the following diagram, in which it will be seen that each successive number is a knight's move from the preceding one, and that five of the dogs (1, 5, 10, 15, and 20) never leave their original kennels.

+-----+------+------+------+------+ |1 |2 |3 |4 |5 | | | | | | | | 1 | 18 | 9 | 14 | 5 | | | | | | | +-----+------+------+------+------+ |6 |7 |8 |9 |10 | | | | | | | | 8 | 13 | 4 | 19 | 10 | | | | | | | +-----+------+------+------+------+ |11 |12 |13 |14 |15 | | | | | | | | 17 | 2 | 11 | 6 | 15 | | | | | | | +-----+------+------+------+------+ |16 |17 |18 |19 |20 | | | | | | | | 12 | 7 | 16 | 3 | 20 | | | | | | | +-----+------+------+------+------+ |21 |22 |23 |24 |25 | | | | | | | | | | | | | | | | | | | +-----+------+------+------+------+ [Ill.u.s.tration]

This position may be arrived at in as few as forty-six moves, as follows: 16--21, 16--22, 16--23, 17--16, 12--17, 12--22, 12--21,7--12, 7--17, 7--22, 11--12, 11--17, 2--7, 2--12, 6--11, 8--7, 8--6, 13--8, 18--13, 11--18, 2--17, 18--12, 18--7, 18--2, 13--7, 3--8, 3--13, 4--3, 4--8, 9--4, 9--3, 14--9, 14--4, 19--14, 19--9, 3--14, 3--19, 6--12, 6--13, 6--14, 17--11, 12--16, 2--12, 7--17, 11--13, 16--18 = 46 moves. I am, of course, not able to say positively that a solution cannot be discovered in fewer moves, but I believe it will be found a very hard task to reduce the number.

345.--THE TWO p.a.w.nS.

Call one p.a.w.n A and the other B. Now, owing to that optional first move, either p.a.w.n may make either 5 or 6 moves in reaching the eighth square. There are, therefore, four cases to be considered: (1) A 6 moves and B 6 moves; (2) A 6 moves and B 5 moves; (3) A 5 moves and B 6 moves; (4) A 5 moves and B 5 moves. In case (1) there are 12 moves, and we may select any 6 of these for A. Therefore 789101112 divided by 123456 gives us the number of variations for this case--that is, 924. Similarly for case (2), 6 selections out of 11 will be 462; in case (3), 5 selections out of 11 will also be 462; and in case (4), 5 selections out of 10 will be 252. Add these four numbers together and we get 2,100, which is the correct number of different ways in which the p.a.w.ns may advance under the conditions. (See No. 270, on p. 204.) 346.--SETTING THE BOARD.

The White p.a.w.ns may be arranged in 40,320 ways, the White rooks in 2 ways, the bishops in 2 ways, and the knights in 2 ways. Multiply these numbers together, and we find that the White pieces may be placed in 322,560 different ways. The Black pieces may, of course, be placed in the same number of ways. Therefore the men may be set up in 322,560 322,560 = 104,044,953,600 ways. But the point that nearly everybody overlooks is that the board may be placed in two different ways for every arrangement. Therefore the answer is doubled, and is 208,089,907,200 different ways.

347.--COUNTING THE RECTANGLES.

There are 1,296 different rectangles in all, 204 of which are squares, counting the square board itself as one, and 1,092 rectangles that are not squares. The general formula is that a board of n squares contains ((n + n))/4 rectangles, of which (2n + 3n + n)/6 are squares and (3n^4 + 2n - 3n - 2n)/12 are rectangles that are not squares. It is curious and interesting that the total number of rectangles is always the square of the triangular number whose side is n.

348.--THE ROOKERY.

The answer involves the little point that in the final position the numbered rooks must be in numerical order in the direction contrary to that in which they appear in the original diagram, otherwise it cannot be solved. Play the rooks in the following order of their numbers. As there is never more than one square to which a rook can move (except on the final move), the notation is obvious--5, 6, 7, 5, 6, 4, 3, 6, 4, 7, 5, 4, 7, 3, 6, 7, 3, 5, 4, 3, 1, 8, 3, 4, 5, 6, 7, 1, 8, 2, 1, and rook takes bishop, checkmate. These are the fewest possible moves--thirty-two. The Black king's moves are all forced, and need not be given.

349.--STALEMATE.

Working independently, the same position was arrived at by Messrs. S. Loyd, E.N. Frankenstein, W.H. Thompson, and myself. So the following may be accepted as the best solution possible to this curious problem :-- White. Black. 1. P--Q4 1. P--K4 2. Q--Q3 2. Q--R5 3. Q--KKt3 3. B--Kt5 ch 4. Kt--Q2 4. P--QR4 5. P--R4 5. P--Q3 6. P--R3 6. B--K3 7. R--R3 7. P--KB4 8. Q--R2 8. P--B4 9. R--KKt3 9. B--Kt6 10. P--QB4 10. P--B5 11. P--B3 11. P--K5 12. P--Q5 12. P--K6 And White is stalemated.

We give a diagram of the curious position arrived at. It will be seen that not one of White's pieces may be moved.

[Ill.u.s.tration]

+-+-+-+-+-+-+-+-+ |r|n| | |k| |n|r| +-+-+-+-+-+-+-+-+ | |p| | | | |p|p| +-+-+-+-+-+-+-+-+ | | | |p| | | | | +-+-+-+-+-+-+-+-+ |p| |p|P| | | | | +-+-+-+-+-+-+-+-+ |P|b|P| | |p| |q| +-+-+-+-+-+-+-+-+ | |b| | |p|P|R|P| +-+-+-+-+-+-+-+-+ | |P| |N|P| |P|Q| +-+-+-+-+-+-+-+-+ | | |B| |K|B|N|R| +-+-+-+-+-+-+-+-+ 350.--THE FORSAKEN KING.

Play as follows:-- White. Black. 1. P to K 4th 1. Any move 2. Q to Kt 4th 2. Any move except on KB file (a) 3. Q to Kt 7th 3. K moves to royal row 4. B to Kt 5th 4. Any move 5. Mate in two moves If 3. K other than to royal row 4. P to Q 4th 4. Any move 5. Mate in two moves (a) If 2. Any move on KB file 3. Q to Q 7th 3. K moves to royal row 4. P to Q Kt 3rd 4. Any move 5. Mate in two moves If 3. K other than to royal row 4. P to Q 4th 4. Any move 5. Mate in two moves Of course, by "royal row" is meant the row on which the king originally stands at the beginning of a game. Though, if Black plays badly, he may, in certain positions, be mated in fewer moves, the above provides for every variation he can possibly bring about.

351.--THE CRUSADER.

White. Black. 1. Kt to QB 3rd 1. P to Q 4th 2. Kt takes QP 2. Kt to QB 3rd 3. Kt takes KP 3. P to KKt 4th 4. Kt takes B 4. Kt to KB 3rd 5. Kt takes P 5. Kt to K 5th 6. Kt takes Kt 6. Kt to B 6th 7. Kt takes Q 7. R to KKt sq 8. Kt takes BP 8. R to KKt 3rd 9. Kt takes P 9. R to K 3rd 10. Kt takes P 10. Kt to Kt 8th 11. Kt takes B 11. R to R 6th 12. Kt takes R 12. P to Kt 4th 13. Kt takes P (ch) 13. K to B 2nd 14. Kt takes P 14. K to Kt 3rd 15. Kt takes R 15. K to R 4th 16. Kt takes Kt 16. K to R 5th White now mates in three moves. 17. P to Q 4th 17. K to R 4th 18. Q to Q 3rd 18. K moves 19. Q to KR 3rd (mate) If 17. K to Kt 5th 18. P to K 4th (dis. ch) 18. K moves 19. P to KKt 3rd (mate) The position after the sixteenth move, with the mate in three moves, was first given by S. Loyd in Chess Nuts.

352.--IMMOVABLE p.a.w.nS.

1. Kt to KB 3 2. Kt to KR 4 3. Kt to Kt 6 4. Kt takes R 5. Kt to Kt 6 6. Kt takes B 7. K takes Kt 8. Kt to QB 3 9. Kt to R 4 10. Kt to Kt 6 11. Kt takes R 12. Kt to Kt 6 13. Kt takes B 14. Kt to Q 6 15. Q to K sq 16. Kt takes Q 17. K takes Kt, and the position is reached.

Black plays precisely the same moves as White, and therefore we give one set of moves only. The above seventeen moves are the fewest possible.

353.--THIRTY-SIX MATES.

Place the remaining eight White pieces thus: K at KB 4th, Q at QKt 6th, R at Q 6th, R at KKt 7th, B at Q 5th, B at KR 8th, Kt at QR 5th, and Kt at QB 5th. The following mates can then be given:-- By discovery from Q 8 By discovery from R at Q 6th 13 By discovery from B at R 8th 11 Given by Kt at R 5th 2 Given by p.a.w.ns 2 -- Total 36 Is it possible to construct a position in which more than thirty-six different mates on the move can be given? So far as I know, n.o.body has yet beaten my arrangement.

354.--AN AMAZING DILEMMA.

Mr Black left his king on his queen's knight's 7th, and no matter what piece White chooses for his p.a.w.n, Black cannot be checkmated. As we said, the Black king takes no notice of checks and never moves. White may queen his p.a.w.n, capture the Black rook, and bring his three pieces up to the attack, but mate is quite impossible. The Black king cannot be left on any other square without a checkmate being possible.

The late Sam Loyd first pointed out the peculiarity on which this puzzle is based.

355.--CHECKMATE!

Remove the White p.a.w.n from B 6th to K 4th and place a Black p.a.w.n on Black's KB 2nd. Now, White plays P to K 5th, check, and Black must play P to B 4th. Then White plays P takes P en pa.s.sant, checkmate. This was therefore White's last move, and leaves the position given. It is the only possible solution.

356.--QUEER CHESS.

+-+-+-+-+-+-+-+-+ | | | | | | | | | +-+-+-+-+-+-+-+-+ | | |R|k|R|N| | | +-+-+-+-+-+-+-+-+ | | | | | | | | | +-+-+-+-+-+-+-+-+ If you place the pieces as follows (where only a portion of the board is given, to save s.p.a.ce), the Black king is in check, with no possible move open to him. The reader will now see why I avoided the term "checkmate," apart from the fact that there is no White king. The position is impossible in the game of chess, because Black could not be given check by both rooks at the same time, nor could he have moved into check on his last move.

I believe the position was first published by the late S. Loyd.

357.--ANCIENT CHINESE PUZZLE.

Play as follows:-- 1. R--Q 6 2. K--R 7 3. R (R 6)--B 6 (mate).

Black's moves are forced, so need not be given.

358.--THE SIX p.a.w.nS.

The general formula for six p.a.w.ns on all squares greater than 2 is this: Six times the square of the number of combinations of n things taken three at a time, where n represents the number of squares on the side of the board. Of course, where n is even the unoccupied squares in the rows and columns will be even, and where n is odd the number of squares will be odd. Here n is 8, so the answer is 18,816 different ways. This is "The Dyer's Puzzle" (Canterbury Puzzles, No. 27) in another form. I repeat it here in order to explain a method of solving that will be readily grasped by the novice. First of all, it is evident that if we put a p.a.w.n on any line, we must put a second one in that line in order that the remainder may be even in number. We cannot put four or six in any row without making it impossible to get an even number in all the columns interfered with. We have, therefore, to put two p.a.w.ns in each of three rows and in each of three columns. Now, there are just six schemes or arrangements that fulfil these conditions, and these are shown in Diagrams A to F, inclusive, on next page.

[Ill.u.s.tration]

I will just remark in pa.s.sing that A and B are the only distinctive arrangements, because, if you give A a quarter-turn, you get F; and if you give B three quarter-turns in the direction that a clock hand moves, you will get successively C, D, and E. No matter how you may place your six p.a.w.ns, if you have complied with the conditions of the puzzle they will fall under one of these arrangements. Of course it will be understood that mere expansions do not destroy the essential character of the arrangements. Thus G is only an expansion of form A. The solution therefore consists in finding the number of these expansions. Supposing we confine our operations to the first three rows, as in G, then with the pairs a and b placed in the first and second columns the pair c may be disposed in any one of the remaining six columns, and so give six solutions. Now slide pair b into the third column, and there are five possible positions for c. Slide b into the fourth column, and c may produce four new solutions. And so on, until (still leaving a in the first column) you have b in the seventh column, and there is only one place for c--in the eighth column. Then you may put a in the second column, b in the third, and c in the fourth, and start sliding c and b as before for another series of solutions.

We find thus that, by using form A alone and confining our operations to the three top rows, we get as many answers as there are combinations of 8 things taken 3 at a time. This is (8 7 6)/(1 2 3) = 56. And it will at once strike the reader that if there are 56 different ways of electing the columns, there must be for each of these ways just 56 ways of selecting the rows, for we may simultaneously work that "sliding" process downwards to the very bottom in exactly the same way as we have worked from left to right. Therefore the total number of ways in which form A may be applied is 56 6 = 3,136. But there are, as we have seen, six arrangements, and we have only dealt with one of these, A. We must, therefore, multiply this result by 6, which gives us 3,136 6 = 18,816, which is the total number of ways, as we have already stated.

359.--COUNTER SOLITAIRE.

Play as follows: 3--11, 9--10, 1--2, 7--15, 8--16, 8--7, 5--13, 1--4, 8--5, 6--14, 3--8, 6--3, 6--12, 1--6, 1--9, and all the counters will have been removed, with the exception of No. 1, as required by the conditions.

360.--CHESSBOARD SOLITAIRE.

Play as follows: 7--15, 8--16, 8--7, 2--10, 1--9, 1--2, 5--13, 3--4, 6--3, 11--1, 14--8, 6--12, 5--6, 5--11, 31--23, 32--24, 32--31, 26--18, 25--17, 25--26, 22--32, 14--22, 29--21, 14--29, 27--28, 30--27, 25--14, 30--20, 25--30, 25--5. The two counters left on the board are 25 and 19--both belonging to the same group, as stipulated--and 19 has never been moved from its original place.

I do not think any solution is possible in which only one counter is left on the board.

361.--THE MONSTROSITY.

White Black, 1. P to KB 4 P to QB 3 2. K to B 2 Q to R 4 3. K to K 3 K to Q sq 4. P to B 5 K to B 2 5. Q to K sq K to Kt 3 6. Q to Kt 3 Kt to QR 3 7. Q to Kt 8 P to KR 4 8. Kt to KB 3 R to R 3 9. Kt to K 5 R to Kt 3 10. Q takes B R to Kt 6, ch 11. P takes R K to Kt 4 12. R to R 4 P to B 3 13. R to Q 4 P takes Kt 14. P to QKt 4 P takes R, ch 15. K to B 4 P to R 5 16. Q to K 8 P to R 6 17. Kt to B 3, ch P takes Kt 18. B to R 3 P to R 7 19. R to Kt sq P to R 8 (Q) 20. R to Kt 2 P takes R 21. K to Kt 5 Q to KKt 8 22. Q to R 5 K to R 5 23. P to Kt 5 R to B sq 24. P to Kt 6 R to B 2 25. P takes R P to Kt 8 (B) 26. P to B 8 (R) Q to B 2 27. B to Q 6 Kt to Kt 5 28. K to Kt 6 K to R 6 29. R to R 8 K to Kt 7 30. P to R 4 Q (Kt 8) to Kt 3 31. P to R 5 K to B 8 32. P takes Q K to Q 8 33. P takes Q K to K 8 34. K to B 7 Kt to KR 3, ch 35. K to K 8 B to R 7 36. P to B 6 B to Kt sq 37. P to B 7 K takes B 38. P to B 8 (B) Kt to Q 4 39. B to Kt 8 Kt to B 3, ch 40. K to Q 8 Kt to K sq 41. P takes Kt (R) Kt to B 2, ch 42. K to B 7 Kt to Q sq 43. Q to B 7, ch K to Kt 8 And the position is reached.

The order of the moves is immaterial, and this order may be greatly varied. But, although many attempts have been made, n.o.body has succeeded in reducing the number of my moves.

362.--THE Wa.s.sAIL BOWL.

The division of the twelve pints of ale can be made in eleven manipulations, as below. The six columns show at a glance the quant.i.ty of ale in the barrel, the five-pint jug, the three-pint jug, and the tramps X, Y, and Z respectively after each manipulation.

Barrel. 5-pint. 3-pint. X. Y. Z.

7 .. 5 .. 0 .. 0 .. 0 .. 0 7 .. 2 .. 3 .. 0 .. 0 .. 0 7 .. 0 .. 3 .. 2 .. 0 .. 0 7 .. 3 .. 0 .. 2 .. 0 .. 0 4 .. 3 .. 3 .. 2 .. 0 .. 0 0 .. 3 .. 3 .. 2 .. 4 .. 0 0 .. 5 .. 1 .. 2 .. 4 .. 0 0 .. 5 .. 0 .. 2 .. 4 .. 1 0 .. 2 .. 3 .. 2 .. 4 .. 1 0 .. 0 .. 3 .. 4 .. 4 .. 1 0 .. 0 .. 0 .. 4 .. 4 .. 4 And each man has received his four pints of ale.

363.--THE DOCTOR'S QUERY.

The mixture of spirits of wine and water is in the proportion of 40 to 1, just as in the other bottle it was in the proportion of 1 to 40.

364.--THE BARREL PUZZLE.

[Ill.u.s.tration: Figs. 1, 2, and 3]

All that is necessary is to tilt the barrel as in Fig. 1, and if the edge of the surface of the water exactly touches the lip a at the same time that it touches the edge of the bottom b, it will be just half full. To be more exact, if the bottom is an inch or so from the ground, then we can allow for that, and the thickness of the bottom, at the top. If when the surface of the water reached the lip a it had risen to the point c in Fig. 2, then it would be more than half full. If, as in Fig. 3, some portion of the bottom were visible and the level of the water fell to the point d, then it would be less than half full.

This method applies to all symmetrically constructed vessels.

365.--NEW MEASURING PUZZLE.

The following solution in eleven manipulations shows the contents of every vessel at the start and after every manipulation:-- 10-quart. 10-quart. 5-quart. 4-quart.

10 .. 10 .. 0 .. 0 5 .. 10 .. 5 .. 0 5 .. 10 .. 1 .. 4 9 .. 10 .. 1 .. 0 9 .. 6 .. 1 .. 4 9 .. 7 .. 0 .. 4 9 .. 7 .. 4 .. 0 9 .. 3 .. 4 .. 4 9 .. 3 .. 5 .. 3 9 .. 8 .. 0 .. 3 4 .. 8 .. 5 .. 3 4 .. 10 .. 3 .. 3 366.--THE HONEST DAIRYMAN.

Whatever the respective quant.i.ties of milk and water, the relative proportion sent to London would always be three parts of water to one of milk. But there are one or two points to be observed. There must originally be more water than milk, or there will be no water in A to double in the second transaction. And the water must not be more than three times the quant.i.ty of milk, or there will not be enough liquid in B to effect the second transaction. The third transaction has no effect on A, as the relative proportions in it must be the same as after the second transaction. It was introduced to prevent a quibble if the quant.i.ty of milk and water were originally the same; for though double "nothing" would be "nothing," yet the third transaction in such a case could not take place.

367.--WINE AND WATER.

The wine in small gla.s.s was one-sixth of the total liquid, and the wine in large gla.s.s two-ninths of total. Add these together, and we find that the wine was seven-eighteenths of total fluid, and therefore the water eleven-eighteenths.

368.--THE KEG OF WINE.

The capacity of the jug must have been a little less than three gallons. To be more exact, it was 2.93 gallons.

369.--MIXING THE TEA.

There are three ways of mixing the teas. Taking them in the order of quality, 2s. 6d., 2s. 3d., 1s. 9p., mix 16 lbs., 1 lb., 3 lbs.; or 14 lbs., 4 lbs., 2 lbs.; or 12 lbs., 7 lbs., 1 lb. In every case the twenty pounds mixture should be worth 2s. 4d. per pound; but the last case requires the smallest quant.i.ty of the best tea, therefore it is the correct answer.

370.--A PACKING PUZZLE.

On the side of the box, 14 by 22+4/5, we can arrange 13 rows containing alternately 7 and 6 b.a.l.l.s, or 85 in all. Above this we can place another layer consisting of 12 rows of 7 and 6 alternately, or a total of 78. In the length of 24+9/10 inches 15 such layers may be packed, the alternate layers containing 85 and 78 b.a.l.l.s. Thus 8 times 85 added to 7 times 78 gives us 1,226 for the full contents of the box.