Part 38
The correct solution to this puzzle is shown in the diagram by the dark line. The five moves indicated will take the queen the greatest distance that it is possible for her to go in five moves, within the conditions. The dotted line shows the route that most people suggest, but it is not quite so long as the other. Let us assume that the distance from the centre of any square to the centre of the next in the same horizontal or vertical line is 2 inches, and that the queen travels from the centre of her original square to the centre of the one at which she rests. Then the first route will be found to exceed 67.9 inches, while the dotted route is less than 67.8 inches. The difference is small, but it is sufficient to settle the point as to the longer route. All other routes are shorter still than these two.
[Illustration]
334.--ST. GEORGE AND THE DRAGON.
We select for the solution of this puzzle one of the prettiest designs that can be formed by representing the moves of the knight by lines from square to square. The chequering of the squares is omitted to give greater clearness. St. George thus slays the Dragon in strict accordance with the conditions and in the elegant manner we should expect of him.
[Illustration: St. George and the Dragon.]
335.--FARMER LAWRENCE'S CORNFIELDS.
There are numerous solutions to this little agricultural problem. The version I give in the next column is rather curious on account of the long parallel straight lines formed by some of the moves.
[Illustration: Farmer Lawrence's Cornfields.]
336.--THE GREYHOUND PUZZLE.
There are several interesting points involved in this question. In the first place, if we had made no stipulation as to the positions of the two ends of the string, it is quite impossible to form any such string unless we begin and end in the top and bottom row of kennels. We may begin in the top row and end in the bottom (or, of course, the reverse), or we may begin in one of these rows and end in the same. But we can never begin or end in one of the two central rows. Our places of starting and ending, however, were fixed for us. Yet the first half of our route must be confined entirely to those squares that are distinguished in the following diagram by circles, and the second half will therefore be confined to the squares that are not circled. The squares reserved for the two half-strings will be seen to be symmetrical and similar.
The next point is that the first half-string must end in one of the central rows, and the second half-string must begin in one of these rows. This is now obvious, because they have to link together to form the complete string, and every square on an outside row is connected by a knight's move with similar squares only--that is, circled or non-circled as the case may be. The half-strings can, therefore, only be linked in the two central rows.
[Illustration]
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 illustration 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.
[Illustration]
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.
[Illustration]
In attempting to solve this problem it is first necessary to take the two distinctive compartments of twenty and twelve squares respectively and analyse 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 passing 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.
[Illustration]
It will be seen in the illustration 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.
[Illustration]
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 passing 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 "buttons and string" method of solving this class 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 buttons 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 relationship between the buttons 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.
[Illustrations: 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 Petites 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 "buttons 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|**| +--+--+--+--+
[Illustration: 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 | | | | | | | | | | | | | | | | | | | +-----+------+------+------+------+
[Illustration]
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 PAWNS.
Call one pawn A and the other B. Now, owing to that optional first move, either pawn 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 7×8×9×10×11×12 divided by 1×2×3×4×5×6 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 pawns may advance under the conditions. (See No. 270, on p. 204.)
346.--SETTING THE BOARD.
The White pawns 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 n² squares contains ((n² + n)²)/4 rectangles, of which (2n³ + 3n² + n)/6 are squares and (3n^4 + 2n³ - 3n² - 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.
[Illustration]
+-+-+-+-+-+-+-+-+ |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 PAWNS.