Chapter 34

+===+===I===I===+===I===+===+===+ | ::::| |:::: |:::| ::::| I...+===I...I=======I...I===+...I |:::| |:::: |:::: |:::| | I...I===I===============I===I...I | |:::: ::::| ::::: |:::| I===I=======I=======I=======I===I |:::| ::::| ::::| ::::| | I...I===+...I...+...I...+===+...I | ::::| |:::: |:::| ::::| I...+===I...+===I===+...I===+...I |:::| |:::: |:::: |:::| | I===I...+=======I=======+...I===I | |:::: ::::| ::::: |:::| I...+=======+...I...+=======+...I |:::: ::::| |:::| ::::: | +===+===+===+===+===+===+===+===+

Eighteen is the maximum number of pieces. I give two solutions. The numbered diagram is so cut that the eighteenth piece has the largest area--eight squares--that is possible under the conditions. The second diagram was prepared under the added condition that no piece should contain more than five squares.

No. 74 in _The Canterbury Puzzles_ shows how to cut the board into twelve pieces, all different, each containing five squares, with one square piece of four squares.

294.--THE CHESSBOARD SENTENCE.

+===I===I===I===I=======I=======+ | |:::| |:::| ::::| ::::| I===I...I===I...I...+===I...+===I |:::| ::::: |:::| ::::: | |...|...+===I...I...+===+...+===I | |:::| |:::| ::::| ::::| |...+===+...+===I===I===I=======I |:::: ::::: |:::| ::::: | I===========I===I...I===I===+...| | ::::: |:::| |:::| |:::| |...+===+...|...|...|...I===+...| |:::| |:::| |:::| |:::: | |...|...|...|...I===+...+===+...| | |:::| |:::| ::::: |:::| I===+...+===I...+=======I===+...| |:::: ::::| ::::: |:::: | +===========I===================+

The pieces may be fitted together, as shown in the illustration, to form a perfect chessboard.

295.--THE EIGHT ROOKS.

Obviously there must be a rook in every row and every column. Starting with the top row, it is clear that we may put our first rook on any one of eight different squares. Wherever it is placed, we have the option of seven squares for the second rook in the second row. Then we have six squares from which to select the third row, five in the fourth, and so on. Therefore the number of our different ways must be 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40,320 (that is 8!), which is the correct answer.

How many ways there are if mere reversals and reflections are not counted as different has not yet been determined; it is a difficult problem. But this point, on a smaller square, is considered in the next puzzle.

296.--THE FOUR LIONS.

There are only seven different ways under the conditions. They are as follows: 1 2 3 4, 1 2 4 3, 1 3 2 4, 1 3 4 2, 1 4 3 2, 2 1 4 3, 2 4 1 3. Taking the last example, this notation means that we place a lion in the second square of first row, fourth square of second row, first square of third row, and third square of fourth row. The first example is, of course, the one we gave when setting the puzzle.

297.--BISHOPS--UNGUARDED.

+...+...+...+...+...+...+...+...+ : ::::: ::::: ::::: ::::: +...+...+...+...+...+...+...+...+ ::::: ::::: ::::: ::::: : +...+...+...+...+...+...+...+...+ : ::::: ::::: ::::: ::::: +...+...+...+...+...+...+...+...+ ::B:: B ::B:: B ::B:: B ::B:: B : +...+...+...+...+...+...+...+...+ : ::::: ::::: ::::: ::::: +...+...+...+...+...+...+...+...+ ::::: ::::: ::::: ::::: : +...+...+...+...+...+...+...+...+ ::::: ::::: ::::: ::::: : +...+...+...+...+...+...+...+...+ : ::::: ::::: ::::: ::::: +...+...+...+...+...+...+...+...+

This cannot be done with fewer bishops than eight, and the simplest solution is to place the bishops in line along the fourth or fifth row of the board (see diagram). But it will be noticed that no bishop is here guarded by another, so we consider that point in the next puzzle.

298.--BISHOPS--GUARDED.

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

This puzzle is quite easy if you first of all give it a little thought. You need only consider squares of one colour, for whatever can be done in the case of the white squares can always be repeated on the black, and they are here quite independent of one another. This equality, of course, is in consequence of the fact that the number of squares on an ordinary chessboard, sixty-four, is an even number. If a square chequered board has an odd number of squares, then there will always be one more square of one colour than of the other.

Ten bishops are necessary in order that every square shall be attacked and every bishop guarded by another bishop. I give one way of arranging them in the diagram. It will be noticed that the two central bishops in the group of six on the left-hand side of the board serve no purpose, except to protect those bishops that are on adjoining squares. Another solution would therefore be obtained by simply raising the upper one of these one square and placing the other a square lower down.

299.--BISHOPS IN CONVOCATION.

The fourteen bishops may be placed in 256 different ways. But every bishop must always be placed on one of the sides of the board--that is, somewhere on a row or file on the extreme edge. The puzzle, therefore, consists in counting the number of different ways that we can arrange the fourteen round the edge of the board without attack. This is not a difficult matter. On a chessboard of n² squares 2n - 2 bishops (the maximum number) may always be placed in 2^n ways without attacking. On an ordinary chessboard n would be 8; therefore 14 bishops may be placed in 256 different ways. It is rather curious that the general result should come out in so simple a form.

300.--THE EIGHT QUEENS.

The solution to this puzzle is shown in the diagram. It will be found that no queen attacks another, and also that no three queens are in a straight line in any oblique direction. This is the only arrangement out of the twelve fundamentally different ways of placing eight queens without attack that fulfils the last condition.

301.--THE EIGHT STARS.

The solution of this puzzle is shown in the first diagram. It is the only possible solution within the conditions stated. But if one of the eight stars had not already been placed as shown, there would then have been eight ways of arranging the stars according to this scheme, if we count reversals and reflections as different. If you turn this page round so that each side is in turn at the bottom, you will get the four reversals; and if you reflect each of these in a mirror, you will get the four reflections. These are, therefore, merely eight aspects of one "fundamental solution." But without that first star being so placed, there is another fundamental solution, as shown in the second diagram. But this arrangement being in a way symmetrical, only produces four different aspects by reversal and reflection.

302.--A PROBLEM IN MOSAICS.

The diagram shows how the tiles may be rearranged. As before, one yellow and one purple tile are dispensed with. I will here point out that in the previous arrangement the yellow and purple tiles in the seventh row might have changed places, but no other arrangement was possible.

303.--UNDER THE VEIL.

Some schemes give more diagonal readings of four letters than others, and we are at first tempted to favour these; but this is a false scent, because what you appear to gain in this direction you lose in others. Of course it immediately occurs to the solver that every LIVE or EVIL is worth twice as much as any other word, since it reads both ways and always counts as 2. This is an important consideration, though sometimes those arrangements that contain most readings of these two words are fruitless in other words, and we lose in the general count.

The above diagram is in accordance with the conditions requiring no letter to be in line with another similar letter, and it gives twenty readings of the five words--six horizontally, six vertically, four in the diagonals indicated by the arrows on the left, and four in the diagonals indicated by the arrows on the right. This is the maximum.

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.

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.

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.

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.:--

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.

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.

312.--THE FIVE CRESCENTS OF BYZANTIUM.

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.

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.)

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.)

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.

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.

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.

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.

The smallest possible number of knights with which this puzzle can be solved is fourteen.