Part 37
Four sets of eight letters may be placed on the board of sixty-four squares in as many as 604 different ways, without any letter ever being in line with a similar one. This does not count reversals and reflections as different, and it does not take into consideration the actual permutations of the letters among themselves; that is, for example, making the L's change places with the E's. Now it is a singular fact that not only do the twenty word-readings that I have given prove to be the real maximum, but there is actually only that one arrangement from which this maximum may be obtained. But if you make the V's change places with the I's, and the L's with the E's, in the solution given, you still get twenty readings--the same number as before in every direction. Therefore there are two ways of getting the maximum from the same arrangement. The minimum number of readings is zero--that is, the letters can be so arranged that no word can be read in any of the directions.
304.--BACHET'S SQUARE.
[Illustration: 1]
[Illustration: 2]
[Illustration: 3]
[Illustration: 4]
Let us use the letters A, K, Q, J, to denote ace, king, queen, jack; and D, S, H, C, to denote diamonds, spades, hearts, clubs. In Diagrams 1 and 2 we have the two available ways of arranging either group of letters so that no two similar letters shall be in line--though a quarter-turn of 1 will give us the arrangement in 2. If we superimpose or combine these two squares, we get the arrangement of Diagram 3, which is one solution. But in each square we may put the letters in the top line in twenty-four different ways without altering the scheme of arrangement. Thus, in Diagram 4 the S's are similarly placed to the D's in 2, the H's to the S's, the C's to the H's, and the D's to the C's. It clearly follows that there must be 24×24 = 576 ways of combining the two primitive arrangements. But the error that Labosne fell into was that of assuming that the A, K, Q, J must be arranged in the form 1, and the D, S, H, C in the form 2. He thus included reflections and half-turns, but not quarter-turns. They may obviously be interchanged. So that the correct answer is 2 × 576 = 1,152, counting reflections and reversals as different. Put in another manner, the pairs in the top row may be written in 16 × 9 × 4 × 1 = 576 different ways, and the square then completed in 2 ways, making 1,152 ways in all.
305.--THE THIRTY-SIX LETTER BLOCKS.
I pointed out that it was impossible to get all the letters into the box under the conditions, but the puzzle was to place as many as possible.
This requires a little judgment and careful investigation, or we are liable to jump to the hasty conclusion that the proper way to solve the puzzle must be first to place all six of one letter, then all six of another letter, and so on. As there is only one scheme (with its reversals) for placing six similar letters so that no two shall be in a line in any direction, the reader will find that after he has placed four different kinds of letters, six times each, every place is occupied except those twelve that form the two long diagonals. He is, therefore, unable to place more than two each of his last two letters, and there are eight blanks left. I give such an arrangement in Diagram 1.
[Illustration: 1]
[Illustration: 2]
The secret, however, consists in not trying thus to place all six of each letter. It will be found that if we content ourselves with placing only five of each letter, this number (thirty in all) may be got into the box, and there will be only six blanks. But the correct solution is to place six of each of two letters and five of each of the remaining four. An examination of Diagram 2 will show that there are six each of C and D, and five each of A, B, E, and F. There are, therefore, only four blanks left, and no letter is in line with a similar letter in any direction.
306.--THE CROWDED CHESSBOARD.
[Illustration]
Here is the solution. Only 8 queens or 8 rooks can be placed on the board without attack, while the greatest number of bishops is 14, and of knights 32. But as all these knights must be placed on squares of the same colour, while the queens occupy four of each colour and the bishops 7 of each colour, it follows that only 21 knights can be placed on the same colour in this puzzle. More than 21 knights can be placed alone on the board if we use both colours, but I have not succeeded in placing more than 21 on the "crowded chessboard." I believe the above solution contains the maximum number of pieces, but possibly some ingenious reader may succeed in getting in another knight.
307.--THE COLOURED COUNTERS.
The counters may be arranged in this order:--
R1, B2, Y3, O4, GS. Y4, O5, G1, R2, B3. G2, R3, B4, Y5, O1. B5, Y1, O2, G3, R4. O3, G4, R5, B1, Y2.
308.--THE GENTLE ART OF STAMP-LICKING.
The following arrangement shows how sixteen stamps may be stuck on the card, under the conditions, of a total value of fifty pence, or 4s. 2d.:--
[Illustration]
If, after placing the four 5d. stamps, the reader is tempted to place four 4d. stamps also, he can afterwards only place two of each of the three other denominations, thus losing two spaces and counting no more than forty-eight pence, or 4s. This is the pitfall that was hinted at. (Compare with No. 43, _Canterbury Puzzles_.)
309.--THE FORTY-NINE COUNTERS.
The counters may be arranged in this order:--
A1, B2, C3, D4, E5, F6, G7. F4, G5, A6, B7, C1, D2, E3. D7, E1, F2, G3, A4, B5, C6. B3, C4, D5, E6, F7, G1, A2. G6, A7, B1, C2, D3, E4, F5. E2, F3, G4, A5, B6, C7, D1. C5, D6, E7, F1, G2, A3, B4.
310.--THE THREE SHEEP.
The number of different ways in which the three sheep may be placed so that every pen shall always be either occupied or in line with at least one sheep is forty-seven.
The following table, if used with the key in Diagram 1, will enable the reader to place them in all these ways:--
+------------+---------------------------+----------+ | | | No. of | | Two Sheep. | Third Sheep. | Ways. | +------------+---------------------------+----------+ | A and B | C, E, G, K, L, N, or P | 7 | | A and C | I, J, K, or O | 4 | | A and D | M, N, or J | 3 | | A and F | J, K, L, or P | 4 | | A and G | H, J, K, N, O, or P | 6 | | A and H | K, L, N, or O | 4 | | A and O | K or L | 2 | | B and C | N | 1 | | B and E | F, H, K, or L | 4 | | B and F | G, J, N, or O | 4 | | B and G | K, L, or N | 3 | | B and H | J or N | 2 | | B and J | K or L | 2 | | F and G | J | 1 | | | | ---- | | | | 47 | +------------+---------------------------+----------+
This, of course, means that if you place sheep in the pens marked A and B, then there are seven different pens in which you may place the third sheep, giving seven different solutions. It was understood that reversals and reflections do not count as different.
If one pen at least is to be _not_ in line with a sheep, there would be thirty solutions to that problem. If we counted all the reversals and reflections of these 47 and 30 cases respectively as different, their total would be 560, which is the number of different ways in which the sheep may be placed in three pens without any conditions. I will remark that there are three ways in which two sheep may be placed so that every pen is occupied or in line, as in Diagrams 2, 3, and 4, but in every case each sheep is in line with its companion. There are only two ways in which three sheep may be so placed that every pen shall be occupied or in line, but no sheep in line with another. These I show in Diagrams 5 and 6. Finally, there is only one way in which three sheep may be placed so that at least one pen shall not be in line with a sheep and yet no sheep in line with another. Place the sheep in C, E, L. This is practically all there is to be said on this pleasant pastoral subject.
[Illustration]
311.--THE FIVE DOGS PUZZLE.
The diagrams show four fundamentally different solutions. In the case of A we can reverse the order, so that the single dog is in the bottom row and the other four shifted up two squares. Also we may use the next column to the right and both of the two central horizontal rows. Thus A gives 8 solutions. Then B may be reversed and placed in either diagonal, giving 4 solutions. Similarly C will give 4 solutions. The line in D being symmetrical, its reversal will not be different, but it may be disposed in 4 different directions. We thus have in all 20 different solutions.
[Illustration]
312.--THE FIVE CRESCENTS OF BYZANTIUM.
[Illustration]
If that ancient architect had arranged his five crescent tiles in the manner shown in the following diagram, every tile would have been watched over by, or in a line with, at least one crescent, and space would have been reserved for a perfectly square carpet equal in area to exactly half of the pavement. It is a very curious fact that, although there are two or three solutions allowing a carpet to be laid down within the conditions so as to cover an area of nearly twenty-nine of the tiles, this is the only possible solution giving exactly half the area of the pavement, which is the largest space obtainable.
313.--QUEENS AND BISHOP PUZZLE.
[Illustration: FIG. 1.]
[Illustration: FIG. 2.]
The bishop is on the square originally occupied by the rook, and the four queens are so placed that every square is either occupied or attacked by a piece. (Fig. 1.)
I pointed out in 1899 that if four queens are placed as shown in the diagram (Fig. 2), then the fifth queen may be placed on any one of the twelve squares marked a, b, c, d, and e; or a rook on the two squares, c; or a bishop on the eight squares, a, b, and e; or a pawn on the square b; or a king on the four squares, b, c, and e. The only known arrangement for four queens and a knight is that given by Mr. J. Wallis in _The Strand Magazine_ for August 1908, here reproduced. (Fig. 3.)
[Illustration: FIG. 3.]
I have recorded a large number of solutions with four queens and a rook, or bishop, but the only arrangement, I believe, with three queens and two rooks in which all the pieces are guarded is that of which I give an illustration (Fig. 4), first published by Dr. C. Planck. But I have since found the accompanying solution with three queens, a rook, and a bishop, though the pieces do not protect one another. (Fig. 5.)
[Illustration: FIG. 4.]
[Illustration: FIG. 5.]
314.--THE SOUTHERN CROSS.
My readers have been so familiarized with the fact that it requires at least five planets to attack every one of a square arrangement of sixty-four stars that many of them have, perhaps, got to believe that a larger square arrangement of stars must need an increase of planets. It was to correct this possible error of reasoning, and so warn readers against another of those numerous little pitfalls in the world of puzzledom, that I devised this new stellar problem. Let me then state at once that, in the case of a square arrangement of eighty one stars, there are several ways of placing five planets so that every star shall be in line with at least one planet vertically, horizontally, or diagonally. Here is the solution to the "Southern Cross": --
It will be remembered that I said that the five planets in their new positions "will, of course, obscure five other stars in place of those at present covered." This was to exclude an easier solution in which only four planets need be moved.
315.--THE HAT-PEG PUZZLE.
The moves will be made quite clear by a reference to the diagrams, which show the position on the board after each of the four moves. The darts indicate the successive removals that have been made. It will be seen that at every stage all the squares are either attacked or occupied, and that after the fourth move no queen attacks any other. In the case of the last move the queen in the top row might also have been moved one square farther to the left. This is, I believe, the only solution to the puzzle.
[Illustration: 1]
[Illustration: 2]
[Illustration: 3]
[Illustration: 4]
316.--THE AMAZONS.
It will be seen that only three queens have been removed from their positions on the edge of the board, and that, as a consequence, eleven squares (indicated by the black dots) are left unattacked by any queen. I will hazard the statement that eight queens cannot be placed on the chessboard so as to leave more than eleven squares unattacked. It is true that we have no rigid proof of this yet, but I have entirely convinced myself of the truth of the statement. There are at least five different ways of arranging the queens so as to leave eleven squares unattacked.
[Illustration]
317.--A PUZZLE WITH PAWNS.
Sixteen pawns may be placed so that no three shall be in a straight line in any possible direction, as in the diagram. We regard, as the conditions required, the pawns as mere points on a plane.
[Illustration]
318.--LION-HUNTING.
There are 6,480 ways of placing the man and the lion, if there are no restrictions whatever except that they must be on different spots. This is obvious, because the man may be placed on any one of the 81 spots, and in every case there are 80 spots remaining for the lion; therefore 81 × 80 = 6,480. Now, if we deduct the number of ways in which the lion and the man may be placed on the same path, the result must be the number of ways in which they will not be on the same path. The number of ways in which they may be in line is found without much difficulty to be 816. Consequently, 6,480 - 816 = 5,664, the required answer.
The general solution is this: 1/3n(n - 1)(3n² - n + 2). This is, of course, equivalent to saying that if we call the number of squares on the side of a "chessboard" n, then the formula shows the number of ways in which two bishops may be placed without attacking one another. Only in this case we must divide by two, because the two bishops have no distinct individuality, and cannot produce a different solution by mere exchange of places.
319.--THE KNIGHT-GUARDS.
[Illustration: DIAGRAM 1.]
[Illustration: DIAGRAM 2.]
The smallest possible number of knights with which this puzzle can be solved is fourteen.
It has sometimes been assumed that there are a great many different solutions. As a matter of fact, there are only three arrangements--not counting mere reversals and reflections as different. Curiously enough, nobody seems ever to have hit on the following simple proof, or to have thought of dealing with the black and the white squares separately.
[Illustration: DIAGRAM 3.]
[Illustration: DIAGRAM 4.]
[Illustration: DIAGRAM 5.]
Seven knights can be placed on the board on white squares so as to attack every black square in two ways only. These are shown in Diagrams 1 and 2. Note that three knights occupy the same position in both arrangements. It is therefore clear that if we turn the board so that a black square shall be in the top left-hand corner instead of a white, and place the knights in exactly the same positions, we shall have two similar ways of attacking all the white squares. I will assume the reader has made the two last described diagrams on transparent paper, and marked them _1a_ and _2a_. Now, by placing the transparent Diagram _1a_ over 1 you will be able to obtain the solution in Diagram 3, by placing _2a_ over 2 you will get Diagram 4, and by placing _2a_ over 1 you will get Diagram 5. You may now try all possible combinations of those two pairs of diagrams, but you will only get the three arrangements I have given, or their reversals and reflections. Therefore these three solutions are all that exist.
320.--THE ROOK'S TOUR.
[Illustration]
The only possible minimum solutions are shown in the two diagrams, where it will be seen that only sixteen moves are required to perform the feat. Most people find it difficult to reduce the number of moves below seventeen*.
[Illustration: THE ROOK'S TOUR.]
321.--THE ROOK'S JOURNEY.
[Illustration]
I show the route in the diagram. It will be seen that the tenth move lands us at the square marked "10," and that the last move, the twenty-first, brings us to a halt on square "21."
322.--THE LANGUISHING MAIDEN.
The dotted line shows the route in twenty-two straight paths by which the knight may rescue the maiden. It is necessary, after entering the first cell, immediately to return before entering another. Otherwise a solution would not be possible. (See "The Grand Tour," p. 200.)
323.--A DUNGEON PUZZLE.
If the prisoner takes the route shown in the diagram--where for clearness the doorways are omitted--he will succeed in visiting every cell once, and only once, in as many as fifty-seven straight lines. No rook's path over the chessboard can exceed this number of moves.
[Illustration: THE LANGUISHING MAIDEN]
[Illustration: A DUNGEON PUZZLE.]
324.--THE LION AND THE MAN.
First of all, the fewest possible straight lines in each case are twenty-two, and in order that no cell may be visited twice it is absolutely necessary that each should pass into one cell and then immediately "visit" the one from which he started, afterwards proceeding by way of the second available cell. In the following diagram the man's route is indicated by the unbroken lines, and the lion's by the dotted lines. It will be found, if the two routes are followed cell by cell with two pencil points, that the lion and the man never meet. But there was one little point that ought not to be overlooked--"they occasionally got glimpses of one another." Now, if we take one route for the man and merely reverse it for the lion, we invariably find that, going at the same speed, they never get a glimpse of one another. But in our diagram it will be found that the man and the lion are in the cells marked A at the same moment, and may see one another through the open doorways; while the same happens when they are in the two cells marked B, the upper letters indicating the man and the lower the lion. In the first case the lion goes straight for the man, while the man appears to attempt to get in the rear of the lion; in the second case it looks suspiciously like running away from one another!
[Illustration]
325.--AN EPISCOPAL VISITATION.
[Illustration]
In the diagram I show how the bishop may be made to visit every one of his white parishes in seventeen moves. It is obvious that we must start from one corner square and end at the one that is diagonally opposite to it. The puzzle cannot be solved in fewer than seventeen moves.
326.--A NEW COUNTER PUZZLE.
Play as follows: 2--3, 9--4, 10--7, 3--8, 4--2, 7--5, 8--6, 5--10, 6--9, 2--5, 1--6, 6--4, 5--3, 10--8, 4--7, 3--2, 8--1, 7--10. The white counters have now changed places with the red ones, in eighteen moves, without breaking the conditions.
327.--A NEW BISHOP'S PUZZLE.
[Illustration: A]
[Illustration: B]
Play as follows, using the notation indicated by the numbered squares in Diagram A:--
White. | Black. | White. | Black. 1. 18--15 | 1. 3--6 | 10. 20--10 | 10. 1--11 2. 17--8 | 2. 4--13 | 11. 3--9 | 11. 18--12 3. 19--14 | 3. 2--7 | 12. 10--13 | 12. 11--8 4. 15--5 | 4. 6--16 | 13. 19--16 | 13. 2--5 5. 8--3 | 5. 13-18 | 14. 16--1 | 14. 5--20 6. 14--9 | 6. 7--12 | 15. 9--6 | 15. 12--15 7. 5--10 | 7. 16-11 | 16. 13-7 | 16. 8--14 8. 9--19 | 8. 12--2 | 17. 6--3 | 17. 15-18 9. 10--4 | 9. 11-17 | 18. 7--2 | 18. 14--19
Diagram B shows the position after the ninth move. Bishops at 1 and 20 have not yet moved, but 2 and 19 have sallied forth and returned. In the end, 1 and 19, 2 and 20, 3 and 17, and 4 and 18 will have exchanged places. Note the position after the thirteenth move.
328.--THE QUEEN'S TOUR.
[Illustration]
The annexed diagram shows a second way of performing the Queen's Tour. If you break the line at the point J and erase the shorter portion of that line, you will have the required path solution for any J square. If you break the line at I, you will have a non-re-entrant solution starting from any I square. And if you break the line at G, you will have a solution for any G square. The Queen's Tour previously given may be similarly broken at three different places, but I seized the opportunity of exhibiting a second tour.
329.--THE STAR PUZZLE.
The illustration explains itself. The stars are all struck out in fourteen straight strokes, starting and ending at a white star.
[Illustration]
330.--THE YACHT RACE.
The diagram explains itself. The numbers will show the direction of the lines in their proper order, and it will be seen that the seventh course ends at the flag-buoy, as stipulated.
[Illustration]
331.--THE SCIENTIFIC SKATER.
In this case we go beyond the boundary of the square. Apart from that, the moves are all queen moves. There are three or four ways in which it can be done.
Here is one way of performing the feat:--
[Illustration]
It will be seen that the skater strikes out all the stars in one continuous journey of fourteen straight lines, returning to the point from which he started. To follow the skater's course in the diagram it is necessary always to go as far as we can in a straight line before turning.
332.--THE FORTY-NINE STARS.
The illustration shows how all the stars may be struck out in twelve straight strokes, beginning and ending at a black star.
[Illustration]
333.--THE QUEEN'S JOURNEY.