Chapter XVII: Preface: v (16)

The six diagrams on next page show solutions for the cases where we replace 2, 3, 4, 5, 6, and 7 hurdles. The dark lines indicate the hurdles that have been replaced. There are, of course, other ways of making the removals.

276.--THE EIGHT VILLAS.

There are several ways of solving the puzzle, but there is very little difference between them. The solver should, however, first of all bear in mind that in making his calculations he need only consider the four villas that stand at the corners, because the intermediate villas can never vary when the corners are known. One way is to place the numbers nought to 9 one at a time in the top left-hand corner, and then consider each case in turn.

Now, if we place 9 in the corner as shown in the Diagram A, two of the corners cannot be occupied, while the corner that is diagonally opposite may be filled by 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 persons. We thus see that there are 10

+---+---+ +-+-----+ +---+---+
|O OHO O| |OHO O O| |O OHO O|
| H | | + | | +=+ |
|O OHO O| |OHO O O| |O OHOHO|
+-+ +-+-+ +-+-----+ +---+ + |
|O|O O|O| |O|O O O| |O O O|O|
| +---+ | | +-+-+ | | +-+ |
|O O O O| |O O OHO| |O O|O O|
+-------+ +-------+ +-------+
2 3 4

+-----+-+ +-+-----+ +-------+
|O O OHO| |OHO O O| |O O O O|
| +=+ | | +=+ | | +=+=+=+
|O OHO O| |OHOHO O| |OHOHO O|
| +-+-+ + | + +-+ | + + + |
|O|O O|O| |O|O O|O| |O|OHO O|
+=+ +=+ | + +=+ +=+ + |
|O O O O| |OHO O O| |O O|O O|
+-------+ +-+-----+ +---+---+
5 6 7
THE SIXTEEN SHEEP

]

solutions with a 9 in the corner. If, however, we substitute 8, the two corners in the same row and column may contain 0, 0, or 1, 1, or 0, 1, or 1, 0. In the case of B, ten different selections may be made for the fourth corner; but in each of the cases C, D, and E, only nine selections are possible, because we cannot use the 9. Therefore with 8 in the top left-hand corner there are 10 + (3 × 9) = 37 different solutions. If we then try 7 in the corner, the result will be 10 + 27 + 40, or 77 solutions. With 6 we get 10 + 27 + 40 + 49 = 126; with 5, 10 + 27 + 40 + 49 + 54 = 180; with 4, the same as with 5, + 55 = 235 ; with 3, the same as with 4, + 52 = 287; with 2, the same as with 3, + 45 = 332; with 1, the same as with 2, + 34 = 366, and with nought in the top left-hand corner the number of solutions will be found to be 10 + 27 + 40 + 49 + 54 + 55 + 52 + 45 + 34 + 19 = 385. As there is no other number to be placed in the top left-hand corner, we have now only to add these totals together thus, 10 + 37 + 77 + 126 + 180 + 235 + 287 + 332 + 366 + 385 = 2,035. We therefore find that the total number of ways in which tenants may occupy some or all of the eight villas so that there shall be always nine persons living along each side of the square is 2,035. Of course, this method must obviously cover all the reversals and reflections, since each corner in turn is occupied by every number in all possible combinations with the other two corners that are in line with it.

A B C D E
+-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+
|9| |0| |8| |0| |8| |1| |8| |0| |8| |1|
+-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+
| |*| | | |*| | | |*| | | |*| | | |*| |
+-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+
|0| | | |0| | | |1| | | |1| | | |0| | |
+-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+ +-+-+-+

]

Here is a general formula for solving the puzzle: (n² + 3n + 2)(n² + 3n + 3)/6. Whatever may be the stipulated number of residents along each of the sides (which number is represented by n), the total number of different arrangements may be thus ascertained. In our particular case the number of residents was nine. Therefore (81 + 27 + 2) × (81 + 27 + 3) and the product, divided by 6, gives 2,035. If the number of residents had been 0, 1, 2, 3, 4, 5, 6, 7, or 8, the total arrangements would be 1, 7, 26, 70, 155, 301, 532, 876, or 1,365 respectively.

277.--COUNTER CROSSES.

Let us first deal with the Greek Cross. There are just eighteen forms in which the numbers may be paired for the two arms. Here they are:--

12978 13968 14958
34956 24957 23967

23958 13769 14759
14967 24758 23768

12589 23759 13579
34567 14768 24568

14569 23569 14379
23578 14578 25368

15369 24369 23189
24378 15378 45167

24179 25169 34169
35168 34178 25178

Of course, the number in the middle is common to both arms. The first pair is the one I gave as an example. I will suppose that we have written out all these crosses, always placing the first row of a pair in the upright and the second row in the horizontal arm. Now, if we leave the central figure fixed, there are 24 ways in which the numbers in the upright may be varied, for the four counters may be changed in 1 × 2 × 3 × 4 = 24 ways. And as the four in the horizontal may also be changed in 24 ways for every arrangement on the other arm, we find that there are 24 × 24 = 576 variations for every form; therefore, as there are 18 forms, we get 18 × 576 = 10,368 ways. But this will include half the four reversals and half the four reflections that we barred, so we must divide this by 4 to obtain the correct answer to the Greek Cross, which is thus 2,592 different ways. The division is by 4 and not by 8, because we provided against half the reversals and reflections by always reserving one number for the upright and the other for the horizontal.

In the case of the Latin Cross, it is obvious that we have to deal with the same 18 forms of pairing. The total number of different ways in this case is the full number, 18 × 576. Owing to the fact that the upper and lower arms are unequal in length, permutations will repeat by reflection, but not by reversal, for we cannot reverse. Therefore this fact only entails division by 2. But in every pair we may exchange the figures in the upright with those in the horizontal (which we could not do in the case of the Greek Cross, as the arms are there all alike); consequently we must multiply by 2. This multiplication by 2 and division by 2 cancel one another. Hence 10,368 is here the correct answer.

278.--A DORMITORY PUZZLE.

MON. TUES. WED.
+---+---+---+ +---+---+---+ +---+---+---+
| 1 | 2 | 1 | | 1 | 3 | 1 | | 1 | 4 | 1 |
+---+---+---+ +---+---+---+ +---+---+---+
| 2 | | 2 | | 1 | | 1 | | 1 | | 1 |
+---+---+---+ +---+---+---+ +---+---+---+
| 1 | 22| 1 | | 3 | 19| 3 | | 4 | 16| 4 |
+---+---+---+ +---+---+---+ +---+---+---+

THURS. FRI. SAT.
+---+---+---+ +---+---+---+ +---+---+---+
| 1 | 5 | 1 | | 2 | 6 | 2 | | 4 | 4 | 4 |
+---+---+---+ +---+---+---+ +---+---+---+
| 2 | | 2 | | 1 | | 1 | | 4 | | 4 |
+---+---+---+ +---+---+---+ +---+---+---+
| 4 | 13| 4 | | 7 | 6 | 7 | | 4 | 4 | 4 |
+---+---+---+ +---+---+---+ +---+---+---+

]

Arrange the nuns from day to day as shown in the six diagrams. The smallest possible number of nuns would be thirty-two, and the arrangements on the last three days admit of variation.

279.--THE BARRELS OF BALSAM.

This is quite easy to solve for any number of barrels--if you know how. This is the way to do it. There are five barrels in each row Multiply the numbers 1, 2, 3, 4, 5 together; and also multiply 6, 7, 8, 9, 10 together. Divide one result by the other, and we get the number of different combinations or selections of ten things taken five at a time. This is here 252. Now, if we divide this by 6 (1 more than the number in the row) we get 42, which is the correct answer to the puzzle, for there are 42 different ways of arranging the barrels. Try this method of solution in the case of six barrels, three in each row, and you will find the answer is 5 ways. If you check this by trial, you will discover the five arrangements with 123, 124, 125, 134, 135 respectively in the top row, and you will find no others.

The general solution to the problem is, in fact, this:

n
C
2n
-----
n + 1

where 2n equals the number of barrels. The symbol C, of course, implies that we have to find how many combinations, or selections, we can make of 2n things, taken n at a time.

280.--BUILDING THE TETRAHEDRON.

Take your constructed pyramid and hold it so that one stick only lies on the table. Now, four sticks must branch off from it in different directions--two at each end. Any one of the five sticks may be left out of this connection; therefore the four may be selected in 5 different ways. But these four matches may be placed in 24 different orders. And as any match may be joined at either of its ends, they may further be varied (after their situations are settled for any particular arrangement) in 16 different ways. In every arrangement the sixth stick may be added in 2 different ways. Now multiply these results together, and we get 5 × 24 × 16 × 2 = 3,840 as the exact number of ways in which the pyramid may be constructed. This method excludes all possibility of error.

A common cause of error is this. If you calculate your combinations by working upwards from a basic triangle lying on the table, you will get half the correct number of ways, because you overlook the fact that an equal number of pyramids may be built on that triangle downwards, so to speak, through the table. They are, in fact, reflections of the others, and examples from the two sets of pyramids cannot be set up to resemble one another--except under fourth dimensional conditions!

281.--PAINTING A PYRAMID.

It will be convenient to imagine that we are painting our pyramids on the flat cardboard, as in the diagrams, before folding up. Now, if we take any _four_ colours (say red, blue, green, and yellow), they may be applied in only 2 distinctive ways, as shown in Figs, 1 and 2. Any other way will only result in one of these when the pyramids are folded up. If we take any _three_ colours, they may be applied in the 3 ways shown in Figs. 3, 4, and 5. If we take any _two_ colours, they may be applied in the 3 ways shown in Figs. 6, 7, and 8. If we take any _single_ colour, it may obviously be applied in only 1 way. But four colours may be selected in 35 ways out of seven; three in 35 ways; two in 21 ways; and one colour in 7 ways. Therefore 35 applied in 2 ways = 70; 35 in 3 ways = 105; 21 in 3 ways = 63; and 7 in 1 way = 7. Consequently the pyramid may be painted in 245 different ways (70 + 105 + 63 + 7), using the seven colours of the solar spectrum in accordance with the conditions of the puzzle.

1 2
+---------------+ +---------------+
\ R / \ B / \ B / \ R /
\ / \ / \ / \ /
\ / G \ / \ / G \ /
\-------/ \-------/
\ / \ /
\ Y / \ Y /
\ / \ /
' '

3 4 5
+---------------+ +---------------+ +---------------+
\ R / \ R / \ R / \ G / \ Y / \ R /
\ / \ / \ / \ / \ / \ /
\ / G \ / \ / G \ / \ / G \ /
\-------/ \-------/ \-------/
\ / \ / \ /
\ Y / \ Y / \ Y /
\ / \ / \ /
' ' '

6 7 8
+---------------+ +---------------+ +---------------+
\ G / \ Y / \ Y / \ Y / \ G / \ G /
\ / \ / \ / \ / \ / \ /
\ / G \ / \ / G \ / \ / G \ /
\-------/ \-------/ \-------/
\ / \ / \ /
\ Y / \ Y / \ Y /
\ / \ / \ /
' ' '

]

282.--THE ANTIQUARY'S CHAIN.

THE number of ways in which nine things may be arranged in a row without any restrictions is 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 = 362,880. But we are told that the two circular rings must never be together; therefore we must deduct the number of times that this would occur. The number is 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 = 40,320 × 2 = 80,640, because if we consider the two circular links to be inseparably joined together they become as one link, and eight links are capable of 40,320 arrangements; but as these two links may always be put on in the orders AB or BA, we have to double this number, it being a question of arrangement and not of design. The deduction required reduces our total to 282,240. Then one of our links is of a peculiar form, like an 8. We have therefore the option of joining on either one end or the other on every occasion, so we must double the last result. This brings up our total to 564,480.

We now come to the point to which I directed the reader's attention--that every link may be put on in one of two ways. If we join the first finger and thumb of our left hand horizontally, and then link the first finger and thumb of the right hand, we see that the right thumb may be either above or below. But in the case of our chain we must remember that although that 8-shaped link has two independent _ends_ it is like every other link in having only two _sides_--that is, you cannot turn over one end without turning the other at the same time.

We will, for convenience, assume that each link has a black side and a side painted white. Now, if it were stipulated that (with the chain lying on the table, and every successive link falling over its predecessor in the same way, as in the diagram) only the white sides should be uppermost as in A, then the answer would be 564,480, as above--ignoring for the present all reversals of the completed chain. If, however, the first link were allowed to be placed either side up, then we could have either A or B, and the answer would be 2 × 564,480 = 1,128,960; if two links might be placed either way up, the answer would be 4 × 564,480; if three links, then 8 × 564,480, and so on. Since, therefore, every link may be placed either side up, the number will be 564,480 multiplied by 2^9, or by 512. This raises our total to 289,013,760.

But there is still one more point to be considered. We have not yet allowed for the fact that with any given arrangement three of the other arrangements may be obtained by simply turning the chain over through its entire length and by reversing the ends. Thus C is really the same as A, and if we turn this page upside down, then A and C give two other arrangements that are still really identical. Thus to get the correct answer to the puzzle we must divide our last total by 4, when we find that there are just 72,253,440 different ways in which the smith might have put those links together. In other words, if the nine links had originally formed a piece of chain, and it was known that the two circular links were separated, then it would be 72,253,439 chances to 1 that the smith would not have put the links together again precisely as they were arranged before!

283.--THE FIFTEEN DOMINOES.

The reader may have noticed that at each end of the line I give is a four, so that, if we like, we can form a ring instead of a line. It can easily be proved that this must always be so. Every line arrangement will make a circular arrangement if we like to join the ends. Now, curious as it may at first appear, the following diagram exactly represents the conditions when we leave the doubles out of the question and devote our attention to forming circular arrangements. Each number, or half domino, is in line with every other number, so that if we start at any one of the five numbers and go over all the lines of the pentagon once and once only we shall come back to the starting place, and the order of our route will give us one of the circular arrangements for the ten dominoes. Take your pencil and follow out the following route, starting at the 4: 41304210234. You have been over all the lines once only, and by repeating all these figures in this way, 41--13--30--04--42--21--10--02--23--34, you get an arrangement of the dominoes (without the doubles) which will be perfectly clear. Take other routes and you will get other arrangements. If, therefore, we can ascertain just how many of these circular routes are obtainable from the pentagon, then the rest is very easy.

Well, the number of different circular routes over the pentagon is 264. How I arrive at these figures I will not at present explain, because it would take a lot of space. The dominoes may, therefore, be arranged in a circle in just 264 different ways, leaving out the doubles. Now, in any one of these circles the five doubles may be inserted in 2^5 = 32 different ways. Therefore when we include the doubles there are 264 × 32 = 8,448 different circular arrangements. But each of those circles may be broken (so as to form our straight line) in any one of 15 different places. Consequently, 8,448 × 15 gives 126,720 different ways as the correct answer to the puzzle.

-----
| |
/ | | \
/ ----- \
/ . . \
----- . . -----
| | . . | o o |
| o | -.--------.--- | |
| | . . . | o o |
----- . . .. -----
\ . . . . /
----- .. -----
| o | . . |o |
| | --------- | o |
| o |. .| o|
----- -----

]

I purposely refrained from asking the reader to discover in just how many different ways the full set of twenty-eight dominoes may be arranged in a straight line in accordance with the ordinary rules of the game, left to right and right to left of any arrangement counting as different ways. It is an exceedingly difficult problem, but the correct answer is 7,959,229,931,520 ways. The method of solving is very complex.

284.--THE CROSS TARGET.

-- --
(CD)( )
-- --
(AE)(A )
-- -- -- -- -- --
(CE)(E )(A )(AB)(C )(D )
-- -- -- -- -- --
(D )( )(B )(E )(EB)( )
-- -- -- -- -- --
(C )(B )
-- --
( )(ED)
-- --

]

Twenty-one different squares may be selected. Of these nine will be of the size shown by the four A's in the diagram, four of the size shown by the B's, four of the size shown by the C's, two of the size shown by the D's, and two of the size indicated by the upper single A, the upper single E, the lower single C, and the EB. It is an interesting fact that you cannot form any one of these twenty-one squares without using at least one of the six circles marked E.

285.--THE FOUR POSTAGE STAMPS.

Referring to the original diagram, the four stamps may be given in the shape 1, 2, 3, 4, in three ways; in the shape 1, 2, 5, 6, in six ways; in the shape 1, 2, 3, 5, or 1, 2, 3, 7, or 1, 5, 6, 7, or 3, 5, 6, 7, in twenty-eight ways; in shape 1, 2, 3, 6, or 2, 5, 6, 7, in fourteen ways; in shape 1, 2, 6, 7, or 2, 3, 5, 6, or 1, 5, 6, 10, or 2, 5, 6, 9, in fourteen ways. Thus there are sixty-five ways in all.

286.--PAINTING THE DIE.

The 1 can be marked on any one of six different sides. For every side occupied by 1 we have a selection of four sides for the 2. For every situation of the 2 we have two places for the 3. (The 6, 5, and 4 need not be considered, as their positions are determined by the 1, 2, and 3.) Therefore 6, 4, and 2 multiplied together make 48 different ways--the correct answer.

287.--AN ACROSTIC PUZZLE.

There are twenty-six letters in the alphabet, giving 325 different pairs. Every one of these pairs may be reversed, making 650 ways. But every initial letter may be repeated as the final, producing 26 other ways. The total is therefore 676 different pairs. In other words, the answer is the square of the number of letters in the alphabet.

288.--CHEQUERED BOARD DIVISIONS.

There are 255 different ways of cutting the board into two pieces of exactly the same size and shape. Every way must involve one of the five cuts shown in Diagrams A, B, C, D, and E. To avoid repetitions by reversal and reflection, we need only consider cuts that enter at the points a, b, and c. But the exit must always be at a point in a straight line from the entry through the centre. This is the most important condition to remember. In case B you cannot enter at a, or you will get the cut provided for in E. Similarly in C or D, you must not enter the key-line in the same direction as itself, or you will get A or B. If you are working on A or C and entering at a, you must consider joins at one end only of the key-line, or you will get repetitions. In other cases you must consider joins at both ends of the key; but after leaving a in case D, turn always either to right or left--use one direction only. Figs. 1 and 2 are examples under A; 3 and 4 are examples under B; 5 and 6 come under C;

and 7 is a pretty example of D. Of course, E is a peculiar type, and obviously admits of only one way of cutting, for you clearly cannot enter at b or c.

Here is a table of the results:--

a b c Ways.
A = 8 + 17 + 21 = 46
B = 0 + 17 + 21 = 38
C = 15 + 31 + 39 = 85
D = 17 + 29 + 39 = 85
E = 1 + 0 + 0 = 1
-- -- -- ---
41 94 120 255

I have not attempted the task of enumerating the ways of dividing a board 8 × 8--that is, an ordinary chessboard. Whatever the method adopted, the solution would entail considerable labour.

289.--LIONS AND CROWNS.

Here is the solution. It will be seen that each of the four pieces (after making the cuts along the thick lines) is of exactly the same size and shape, and that each piece contains a lion and a crown. Two of the pieces are shaded so as to make the solution quite clear to the eye.

290.--BOARDS WITH AN ODD NUMBER OF SQUARES.

There are fifteen different ways of cutting the 5 × 5 board (with the central square removed) into two pieces of the same size and shape. Limitations of space will not allow me to give diagrams of all these, but I will enable the reader to draw them all out for himself without the slightest difficulty. At whatever point on the edge your cut enters, it must always end at a point on the edge, exactly opposite in a line through the centre of the square. Thus, if you enter at point 1 (see Fig. 1) at the top, you must leave at point 1 at the bottom. Now, 1 and 2 are the only two really different points of entry; if we use any others they will simply produce similar solutions. The directions of the cuts in the following fifteen

solutions are indicated by the numbers on the diagram. The duplication of the numbers can lead to no confusion, since every successive number is contiguous to the previous one. But whichever direction you take from the top downwards you must repeat from the bottom upwards, one direction being an exact reflection of the other.

1, 4, 8.
1, 4, 3, 7, 8.
1, 4, 3, 7, 10, 9.
1, 4, 3, 7, 10, 6, 5, 9.
1, 4, 5, 9.
1, 4, 5, 6, 10, 9.
1, 4, 5, 6, 10, 7, 8.
2, 3, 4, 8.
2, 3, 4, 5, 9.
2, 3, 4, 5, 6, 10, 9.
2, 3, 4, 5, 6, 10, 7, 8.
2, 3, 7, 8.
2, 3, 7, 10, 9.
2, 3, 7, 10, 6, 5, 9.
2, 3, 7, 10, 6, 5, 4, 8.

It will be seen that the fourth direction (1, 4, 3, 7, 10, 6, 5, 9) produces the solution shown in Fig. 2. The thirteenth produces the solution given in propounding the puzzle, where the cut entered at the side instead of at the top. The pieces, however, will be of the same shape if turned over, which, as it was stated in the conditions, would not constitute a different solution.

291.--THE GRAND LAMA'S PROBLEM.

The method of dividing the chessboard so that each of the four parts shall be of exactly the same size and shape, and contain one of the gems, is shown in the diagram. The method of shading the squares is adopted to make the shape of the pieces clear to the eye. Two of the pieces are shaded and two left white.

The reader may find it interesting to compare this puzzle with that of the "Weaver" (No. 14, _Canterbury Puzzles_).

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

]

292.--THE ABBOT'S WINDOW.

THE man who was "learned in strange mysteries" pointed out to Father John that the orders of the Lord Abbot of St. Edmondsbury might be easily carried out by blocking up twelve of the lights in the window as shown by the dark squares in the following sketch:--

+===+===+===+===+===+===+===+===+
| : : : : : : : |
I...+===+...+...+...+...+===+...I
| IIIII : : : IIIII |
I...+===+===+...+...+===+===+...I
| : IIIII : IIIII : |
I...+...+===+===+===+===+...+...I
| : : IIIIIIIII : : |
I...+...+...+===+===+...+...+...I
| : : IIIIIIIII : : |
I...+...+===+===+===+===+...+...I
| : IIIII : IIIII : |
I...+===+===+...+...+===+===+...I
| IIIII : : : IIIII |
I...+===+...+...+...+...+===+...I
| : : : : : : : |
+===+===+===+===+===+===+===+===+

]

Father John held that the four corners should also be darkened, but the sage explained that it was desired to obstruct no more light than was absolutely necessary, and he said, anticipating Lord Dundreary, "A single pane can no more be in a _line_ with itself than one bird can go into a corner and flock in solitude. The Abbot's condition was that no diagonal _lines_ should contain an odd number of lights."

Now, when the holy man saw what had been done he was well pleased, and said, "Truly, Father John, thou art a man of deep wisdom, in that thou hast done that which seemed impossible, and yet withal adorned our window with a device of the cross of St. Andrew, whose name I received from my godfathers and godmothers." Thereafter he slept well and arose refreshed. The window might be seen intact to-day in the monastery of St. Edmondsbury, if it existed, which, alas! the window does not.

293.--THE CHINESE CHESSBOARD.

+===I===+===+===+===I===+===+===+
| |:::: 2 ::::| 3 |:::| 5 |:6:|
I...+===+...+===+...I...I...+===I
|:::: 1 |:::| ::::| 4 |:::| 7 |
I...+===+===+...I===I...I===+===I
| |:::: |:::| ::::| 9 |:::|
I===I...I===============I...I...I
|:::: 11|:::: ::::: 10|:::| 8 |
I=======I===I===========I...I...I
| ::::: 12|:::: 13::::| |:::|
I=======+...I...+===+===|===+===I
|:::: 14|:::| |:::| 16::::| 17|
I...+...I===I===+...+...+===+...I
| ::::| ::::: 15|:::| ::::|
I=======+===========+===+=======I
|:::: ::::: 18::::: ::::: |
+===+===+===+===+===+===+===+===+

+===+===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.

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

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.

_ _ I V E L _ _
E V L _ _ I _ _
L _ _ I _ _ V E
I _ V E _ _ _ L
_ E _ _ L V _ I
_ L I _ _ I _ E V
/V _ E L _ _ I _
_ I _ _ V E L _\

]

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.