Chapter D (1)

Text size

+--+--+--+--+
| 6|11| 4|15|
+--+--+--+--+
| 1|14| 7|10|
+--+--+--+--+
| 8| 5|12| 3|
+--+--+--+--+
|13| 2| 9|**|
+--+--+--+--+

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 |
| | | | | |
| | | | | |
| | | | | |
+-----+------+------+------+------+

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.

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|r|n| | |k| |n|r|
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| |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.

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

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

353.--THIRTY-SIX MATES.

Place the remaining eight White pieces thus: K at KB 4th, Q at QKt 6th, R at Q 6th, R at KKt 7th, B at Q 5th, B at KR 8th, Kt at QR 5th, and Kt at QB 5th. The following mates can then be given:--

By discovery from Q 8
By discovery from R at Q 6th 13
By discovery from B at R 8th 11
Given by Kt at R 5th 2
Given by pawns 2
--
Total 36

Is it possible to construct a position in which more than thirty-six different mates on the move can be given? So far as I know, nobody has yet beaten my arrangement.

354.--AN AMAZING DILEMMA.

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

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

355.--CHECKMATE!

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

356.--QUEER CHESS.

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| | | | | | | | |
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| | |R|k|R|N| | |
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| | | | | | | | |
+-+-+-+-+-+-+-+-+

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

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

357.--ANCIENT CHINESE PUZZLE.

Play as follows:--

1. R--Q 6
2. K--R 7
3. R (R 6)--B 6 (mate).

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

358.--THE SIX PAWNS.

The general formula for six pawns on all squares greater than 2² is this: Six times the square of the number of combinations of n things taken three at a time, where n represents the number of squares on the side of the board. Of course, where n is even the unoccupied squares in the rows and columns will be even, and where n is odd the number of squares will be odd. Here n is 8, so the answer is 18,816 different ways. This is "The Dyer's Puzzle" (_Canterbury Puzzles_, No. 27) in another form. I repeat it here in order to explain a method of solving that will be readily grasped by the novice. First of all, it is evident that if we put a pawn on any line, we must put a second one in that line in order that the remainder may be even in number. We cannot put four or six in any row without making it impossible to get an even number in all the columns interfered with. We have, therefore, to put two pawns in each of three rows and in each of three columns. Now, there are just six schemes or arrangements that fulfil these conditions, and these are shown in Diagrams A to F, inclusive, on next page.

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

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

359.--COUNTER SOLITAIRE.

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

360.--CHESSBOARD SOLITAIRE.

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

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

361.--THE MONSTROSITY.

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

And the position is reached.

The order of the moves is immaterial, and this order may be greatly varied. But, although many attempts have been made, nobody has succeeded in reducing the number of my moves.

362.--THE WASSAIL BOWL.

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

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

7 .. 5 .. 0 .. 0 .. 0 .. 0
7 .. 2 .. 3 .. 0 .. 0 .. 0
7 .. 0 .. 3 .. 2 .. 0 .. 0
7 .. 3 .. 0 .. 2 .. 0 .. 0
4 .. 3 .. 3 .. 2 .. 0 .. 0
0 .. 3 .. 3 .. 2 .. 4 .. 0
0 .. 5 .. 1 .. 2 .. 4 .. 0
0 .. 5 .. 0 .. 2 .. 4 .. 1
0 .. 2 .. 3 .. 2 .. 4 .. 1
0 .. 0 .. 3 .. 4 .. 4 .. 1
0 .. 0 .. 0 .. 4 .. 4 .. 4

And each man has received his four pints of ale.

363.--THE DOCTOR'S QUERY.

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

364.--THE BARREL PUZZLE.

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

This method applies to all symmetrically constructed vessels.

365.--NEW MEASURING PUZZLE.

The following solution in eleven manipulations shows the contents of every vessel at the start and after every manipulation:--

10-quart. 10-quart. 5-quart. 4-quart.

10 .. 10 .. 0 .. 0
5 .. 10 .. 5 .. 0
5 .. 10 .. 1 .. 4
9 .. 10 .. 1 .. 0
9 .. 6 .. 1 .. 4
9 .. 7 .. 0 .. 4
9 .. 7 .. 4 .. 0
9 .. 3 .. 4 .. 4
9 .. 3 .. 5 .. 3
9 .. 8 .. 0 .. 3
4 .. 8 .. 5 .. 3
4 .. 10 .. 3 .. 3

366.--THE HONEST DAIRYMAN.

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

367.--WINE AND WATER.

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

368.--THE KEG OF WINE.

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

369.--MIXING THE TEA.

There are three ways of mixing the teas. Taking them in the order of quality, 2s. 6d., 2s. 3d., 1s. 9p., mix 16 lbs., 1 lb., 3 lbs.; or 14 lbs., 4 lbs., 2 lbs.; or 12 lbs., 7 lbs., 1 lb. In every case the twenty pounds mixture should be worth 2s. 4½d. per pound; but the last case requires the smallest quantity of the best tea, therefore it is the correct answer.

370.--A PACKING PUZZLE.

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

371.--GOLD PACKING IN RUSSIA.

The box should be 100 inches by 100 inches by 11 inches deep, internal dimensions. We can lay flat at the bottom a row of eight slabs, lengthways, end to end, which will just fill one side, and nine of these rows will dispose of seventy-two slabs (all on the bottom), with a space left over on the bottom measuring 100 inches by 1 inch by 1 inch. Now make eleven depths of such seventy-two slabs, and we have packed 792, and have a space 100 inches by 1 inch by 11 inches deep. In this we may exactly pack the remaining eight slabs on edge, end to end.

372.--THE BARRELS OF HONEY.

The only way in which the barrels could be equally divided among the three brothers, so that each should receive his 3½ barrels of honey and his 7 barrels, is as follows:--

Full. Half-full. Empty.
A 3 1 3
B 2 3 2
C 2 3 2

There is one other way in which the division could be made, were it not for the objection that all the brothers made to taking more than four barrels of the same description. Except for this difficulty, they might have given B his quantity in exactly the same way as A above, and then have left C one full barrel, five half-full barrels, and one empty barrel. It will thus be seen that in any case two brothers would have to receive their allowance in the same way.

373.--CROSSING THE STREAM.

First, the two sons cross, and one returns Then the man crosses and the other son returns. Then both sons cross and one returns. Then the lady crosses and the other son returns Then the two sons cross and one of them returns for the dog. Eleven crossings in all.

It would appear that no general rule can be given for solving these river-crossing puzzles. A formula can be found for a particular case (say on No. 375 or 376) that would apply to any number of individuals under the restricted conditions; but it is not of much use, for some little added stipulation will entirely upset it. As in the case of the measuring puzzles, we generally have to rely on individual ingenuity.

374.--CROSSING THE RIVER AXE.

Here is the solution:--

| {J 5) | G T8 3
5 | ( J } | G T8 3
5 | {G 3) | JT8
53 | ( G } | JT8
53 | {J T) | G 8
J 5 | (T 3} | G 8
J 5 | {G 8) | T 3
G 8 | (J 5} | T
G 8 | {J T) | 53
JT8 | ( G } | 53
JT8 | {G 3) | 5
G T8 3 | ( J } | 5
G T8 3 | {J 5) |

G, J, and T stand for Giles, Jasper, and Timothy; and 8, 5, 3, for £800, £500, and £300 respectively. The two side columns represent the left bank and the right bank, and the middle column the river. Thirteen crossings are necessary, and each line shows the position when the boat is in mid-stream during a crossing, the point of the bracket indicating the direction.

It will be found that not only is no person left alone on the land or in the boat with more than his share of the spoil, but that also no two persons are left with more than their joint shares, though this last point was not insisted upon in the conditions.

375.--FIVE JEALOUS HUSBANDS.

It is obvious that there must be an odd number of crossings, and that if the five husbands had not been jealous of one another the party might have all got over in nine crossings. But no wife was to be in the company of a man or men unless her husband was present. This entails two more crossings, eleven in all.

The following shows how it might have been done. The capital letters stand for the husbands, and the small letters for their respective wives. The position of affairs is shown at the start, and after each crossing between the left bank and the right, and the boat is represented by the asterisk. So you can see at a glance that a, b, and c went over at the first crossing, that b and c returned at the second crossing, and so on.

ABCDE abcde *|..|
| |
1. ABCDE de |..|* abc
2. ABCDE bcde *|..| a
3. ABCDE e |..|* abcd
4. ABCDE de *|..| abc
5. DE de |,,|* ABC abc
6. CDE cde *|..| AB ab
7. cde |..|* ABCDE ab
8. bcde *|..| ABCDE a
9. e |..|* ABCDE abcd
10. bc e *|..| ABCDE a d
11. |..|* ABCDE abcde

There is a little subtlety concealed in the words "show the _quickest_ way."

Everybody correctly assumes that, as we are told nothing of the rowing capabilities of the party, we must take it that they all row equally well. But it is obvious that two such persons should row more quickly than one.

Therefore in the second and third crossings two of the ladies should take back the boat to fetch d, not one of them only. This does not affect the number of landings, so no time is lost on that account. A similar opportunity occurs in crossings 10 and 11, where the party again had the option of sending over two ladies or one only.

To those who think they have solved the puzzle in nine crossings I would say that in every case they will find that they are wrong. No such jealous husband would, in the circumstances, send his wife over to the other bank to a man or men, even if she assured him that she was coming back next time in the boat. If readers will have this fact in mind, they will at once discover their errors.

376.--THE FOUR ELOPEMENTS.

If there had been only three couples, the island might have been dispensed with, but with four or more couples it is absolutely necessary in order to cross under the conditions laid down. It can be done in seventeen passages from land to land (though French mathematicians have declared in their books that in such circumstances twenty-four are needed), and it cannot be done in fewer. I will give one way. A, B, C, and D are the young men, and a, b, c, and d are the girls to whom they are respectively engaged. The three columns show the positions of the different individuals on the lawn, the island, and the opposite shore before starting and after each passage, while the asterisk indicates the position of the boat on every occasion.

Having found the fewest possible passages, we should consider two other points in deciding on the "quickest method": Which persons were the most expert in handling the oars, and which method entails the fewest possible delays in getting in and out of the boat? We have no data upon which to decide the first point, though it is probable that, as the boat belonged to the girls' household, they would be capable oarswomen. The other point, however, is important, and in the solution I have given (where the girls do 8-13ths of the rowing and A and D need not row at all) there are only sixteen gettings-in and sixteen gettings-out. A man and a girl are never in the boat together, and no man ever lands on the island. There are other methods that require several more exchanges of places.

377.--STEALING THE CASTLE TREASURE.

Here is the best answer, in eleven manipulations:--

Treasure down.
Boy down--treasure up.
Youth down--boy up.
Treasure down.
Man down--youth and treasure up.
Treasure down.
Boy down--treasure up.
Treasure down.
Youth down--boy up.
Boy down--treasure up.
Treasure down.

378.--DOMINOES IN PROGRESSION.

There are twenty-three different ways. You may start with any domino, except the 4--4 and those that bear a 5 or 6, though only certain initial dominoes may be played either way round. If you are given the common difference and the first domino is played, you have no option as to the other dominoes. Therefore all I need do is to give the initial domino for all the twenty-three ways, and state the common difference. This I will do as follows:--

With a common difference of 1, the first domino may be either of these: 0--0, 0--1, 1--0, 0--2, 1--1, 2--0, 0--3, 1--2, 2--1, 3--0, 0--4, 1--3, 2--2, 3--1, 1--4, 2--3, 3--2, 2--4, 3--3, 3--4. With a difference of 2, the first domino may be 0--0, 0--2, or 0--1. Take the last case of all as an example. Having played the 0--1, and the difference being 2, we are compelled to continue with 1--2, 2--3, 3--4. 4--5, 5--6. There are three dominoes that can never be used at all. These are 0--5, 0--6, and 1--6. If we used a box of dominoes extending to 9--9, there would be forty different ways.

379.--THE FIVE DOMINOES.

There are just ten different ways of arranging the dominoes. Here is one of them:--

(2--0) (0--0) (0--1) (1--4) (4--0).

I will leave my readers to find the remaining nine for themselves.

380.--THE DOMINO FRAME PUZZLE.

+---+-------+-------+-------+-------+-------+-------+-------+
| 2 | 2 | 5 | 5 | 6 | 6 | 6 | 6 | 1 | 1 | | | | | 4 |
| - +-------+-------+-------+-------+-------+-------+---+---+
| 2 | | 4 |
+---+ | - |
| 2 | | 3 |
| - | +---+
| 6 | | 3 |
+---+ T H E | - |
| 6 | | 3 |
| - | +---+
| 3 | | 3 |
+---+ | - |
| 3 | | 1 |
| - | D O M I N O F R A M E +---+
| | | 1 |
+---+ | - |
| | | 1 |
| - | +---+
| 5 | | 1 |
+---+ -S-O-L-U-T-I-O-N- | - |
| 5 | | 4 |
| - | +---+
| 3 | | 4 |
+---+ | - |
| 3 | | 6 |
| - | +---+
| 2 | | 6 |
+---+---+-------+-------+-------+-------+-------+-------+ - |
| 2 | 1 | 1 | 5 | 5 | 5 | 5 | 4 | 4 | 4 | 4 | 2 | 2 | | |
+-------+-------+-------+-------+-------+-------+-------+---+

]

The illustration is a solution. It will be found that all four sides of the frame add up 44. The sum of the pips on all the dominoes is 168, and if we wish to make the sides sum to 44, we must take care that the four corners sum to 8, because these corners are counted twice, and 168 added to 8 will equal 4 times 44, which is necessary. There are many different solutions. Even in the example given certain interchanges are possible to produce different arrangements. For example, on the left-hand side the string of dominoes from 2--2 down to 3--2 may be reversed, or from 2--6 to 3--2, or from 3--0 to 5--3. Also, on the right-hand side we may reverse from 4--3 to 1--4. These changes will not affect the correctness of the solution.

381.--THE CARD FRAME PUZZLE.

The sum of all the pips on the ten cards is 55. Suppose we are trying to get 14 pips on every side. Then 4 times 14 is 56. But each of the four corner cards is added in twice, so that 55 deducted from 56, or 1, must represent the sum of the four corner cards. This is clearly impossible; therefore 14 is also impossible. But suppose we came to trying 18. Then 4 times 18 is 72, and if we deduct 55 we get 17 as the sum of the corners. We need then only try different arrangements with the four corners always summing to 17, and we soon discover the following solution:--

+-------+-------+-------+
| 2 | 10 | 6 |
+---+---+------ +---+---+
| | | |
| 3 | | 7 |
| | | |
+---+ +---+
| | | |
| 8 | | 1 |
| | | |
+---+---+-------+--+----+
| 5 | 9 | 4 |
+-------+-------+-------+

]

The final trials are very limited in number, and must with a little judgment either bring us to a correct solution or satisfy us that a solution is impossible under the conditions we are attempting. The two centre cards on the upright sides can, of course, always be interchanged, but I do not call these different solutions. If you reflect in a mirror you get another arrangement, which also is not considered different. In the answer given, however, we may exchange the 5 with the 8 and the 4 with the 1. This is a different solution. There are two solutions with 18, four with 19, two with 20, and two with 22--ten arrangements in all. Readers may like to find all these for themselves.

382.--THE CROSS OF CARDS.

There are eighteen fundamental arrangements, as follows, where I only give the numbers in the horizontal bar, since the remainder must naturally fall into their places.

5 6 1 7 4 2 4 5 6 8
3 5 1 6 8 3 4 5 6 7
3 4 1 7 8 1 4 7 6 8
2 5 1 7 8 2 3 7 6 8
2 5 3 6 8 2 4 7 5 8
1 5 3 7 8 3 4 9 5 6
2 4 3 7 8 2 4 9 5 7
1 4 5 7 8 1 4 9 6 7
2 3 5 7 8 2 3 9 6 7

It will be noticed that there must always be an odd number in the centre, that there are four ways each of adding up 23, 25, and 27, but only three ways each of summing to 24 and 26.

383.--THE "T" CARD PUZZLE.

If we remove the ace, the remaining cards may he divided into two groups (each adding up alike) in four ways; if we remove 3, there are three ways; if 5, there are four ways; if 7, there are three ways; and if we remove 9, there are four ways of making two equal groups. There are thus eighteen different ways of grouping, and if we take any one of these and keep the odd card (that I have called "removed") at the head of the column, then one set of numbers can be varied in order in twenty-four ways in the column and the other four twenty-four ways in the horizontal, or together they may be varied in 24 × 24 = 576 ways. And as there are eighteen such cases, we multiply this number by 18 and get 10,368, the correct number of ways of placing the cards. As this number includes the reflections, we must divide by 2, but we have also to remember that every horizontal row can change places with a vertical row, necessitating our multiplying by 2; so one operation cancels the other.

384.--CARD TRIANGLES.

The following arrangements of the cards show (1) the smallest possible sum, 17; and (2) the largest possible, 23.

1 7
9 6 4 2
4 8 3 6
3 7 5 2 9 5 1 8

It will be seen that the two cards in the middle of any side may always be interchanged without affecting the conditions. Thus there are eight ways of presenting every fundamental arrangement. The number of fundamentals is eighteen, as follows: two summing to 17, four summing to 19, six summing to 20, four summing to 21, and two summing to 23. These eighteen fundamentals, multiplied by eight (for the reason stated above), give 144 as the total number of different ways of placing the cards.

385.--"STRAND" PATIENCE.

The reader may find a solution quite easy in a little over 200 moves, but, surprising as it may at first appear, not more than 62 moves are required. Here is the play: By "4 C up" I mean a transfer of the 4 of clubs with all the cards that rest on it. 1 D on space, 2 S on space, 3 D on space, 2 S on 3 D, 1 H on 2 S, 2 C on space, 1 D on 2 C, 4 S on space, 3 H on 4 S (9 moves so far), 2 S up on 3 H (3 moves), 5 H and 5 D exchanged, and 4 C on 5 D (6 moves), 3 D on 4 C (1), 6 S (with 5 H) on space (3), 4 C up on 5 H (3), 2 C up on 3 D (3), 7 D on space (1), 6 C up on 7 D (3), 8 S on space (1), 7 H on 8 S (1), 8 C on 9 D (1), 7 H on 8 C (1), 8 S on 9 H (1), 7 H on 8 S (1), 7 D up on 8 C (5), 4 C up on 5 D (9), 6 S up on 7 H (3), 4 S up on 5 H (7) = 62 moves in all. This is my record; perhaps the reader can beat it.

386.--A TRICK WITH DICE.

All you have to do is to deduct 250 from the result given, and the three figures in the answer will be the three points thrown with the dice. Thus, in the throw we gave, the number given would be 386; and when we deduct 250 we get 136, from which we know that the throws were 1, 3, and 6.

The process merely consists in giving 100a + 10b + c + 250, where a, b, and c represent the three throws. The result is obvious.

387.--THE VILLAGE CRICKET MATCH.

| Mr. Dumkins >>-->
|------------------------> |
| <------------------- |
| -------------------> |
1 |<----------------------- |
| |
| <------------------------|
| -------------------> |
| <------------------- |
| ----------------------->|
| <--<< Mr. Podder |

| Mr. Luffey >>-->
|------------------------> |
| <------------------- |
| ----------------------->|
2 | |
|<----------------------- |
| -------------------> |
| <------------------------|
<--<< Mr. Struggles |

]

The diagram No. 1 will show that as neither Mr. Podder nor Mr. Dumkins can ever have been within the crease opposite to that from which he started, Mr. Dumkins would score nothing by his performance. Diagram No. 2 will, however, make it clear that since Mr. Luffey and Mr. Struggles have, notwithstanding their energetic but careless movements, contrived to change places, the manoeuvre must increase Mr. Struggles's total by one run.

388.--SLOW CRICKET.

The captain must have been "not out" and scored 21. Thus:--

2 men (each lbw) 19
4 men (each caught) 17
1 man (run out) 0
3 men (each bowled) 9
1 man (captain--not out) 21
-- --
11 66

The captain thus scored exactly 15 more than the average of the team. The "others" who were bowled could only refer to three men, as the eleventh man would be "not out." The reader can discover for himself why the captain must have been that eleventh man. It would not necessarily follow with any figures.

389.--THE FOOTBALL PLAYERS.

The smallest possible number of men is seven. They could be accounted for in three different ways: 1. Two with both arms sound, one with broken right arm, and four with both arms broken. 2. One with both arms sound, one with broken left arm, two with broken right arm, and three with both arms broken. 3. Two with left arm broken, three with right arm broken, and two with both arms broken. But if every man was injured, the last case is the only one that would apply.

390.--THE HORSE-RACE PUZZLE.

The answer is: £12 on Acorn, £15 on Bluebottle, £20 on Capsule.

391.--THE MOTOR-CAR RACE.

The first point is to appreciate the fact that, in a race round a circular track, there are the same number of cars behind one as there are before. All the others are both behind and before. There were thirteen cars in the race, including Gogglesmith's car. Then one-third of twelve added to three-quarters of twelve will give us thirteen--the correct answer.

392.--THE PEBBLE GAME.

In the case of fifteen pebbles, the first player wins if he first takes two. Then when he holds an odd number and leaves 1, 8, or 9 he wins, and when he holds an even number and leaves 4, 5, or 12 he also wins. He can always do one or other of these things until the end of the game, and so defeat his opponent. In the case of thirteen pebbles the first player must lose if his opponent plays correctly. In fact, the only numbers with which the first player ought to lose are 5 and multiples of 8 added to 5, such as 13, 21, 29, etc.

393.--THE TWO ROOKS.

The second player can always win, but to ensure his doing so he must always place his rook, at the start and on every subsequent move, on the same diagonal as his opponent's rook. He can then force his opponent into a corner and win. Supposing the diagram to represent the positions of the rooks at the start, then, if Black played first, White might have placed his rook at A and won next move. Any square on that diagonal from A to H will win, but the best play is always to restrict the moves of the opposing rook as much as possible. If White played first, then Black should have placed his rook at B (F would not be so good, as it gives White more scope); then if White goes to C, Black moves to D; White to E, Black to F; White to G, Black to C; White to H, Black to I; and Black must win next move. If at any time Black had failed to move on to the same diagonal as White, then White could take Black's diagonal and win.

r: black rook
R: white rook

+-+-+-+-+-+-+-+-+
|r| | | | | | | |
+-+-+-+-+-+-+-+-+
| |A| | | | | | |
+-+-+-+-+-+-+-+-+
| | | | | | | | |
+-+-+-+-+-+-+-+-+
| | | | | | | | |
+-+-+-+-+-+-+-+-+
| | | | |B|D|F| |
+-+-+-+-+-+-+-+-+
| | | | | |R|C|E|
+-+-+-+-+-+-+-+-+
| | | | | | |I|G|
+-+-+-+-+-+-+-+-+
| | | | | | | |H|
+-+-+-+-+-+-+-+-+

THE TWO ROOKS.

394.--PUSS IN THE CORNER.

No matter whether he plays first or second, the player A, who starts the game at 55, must win. Assuming that B adopts the very best lines of play in order to prolong as much as possible his existence, A, if he has first move, can always on his 12th move capture B; and if he has the second move, A can always on his 14th move make the capture. His point is always to get diagonally in line with his opponent, and by going to 33, if he has first move, he prevents B getting diagonally in line with himself. Here are two good games. The number in front of the hyphen is always A's move; that after the hyphen is B's:--

33-8, 32-15, 31-22, 30-21, 29-14, 22-7, 15-6, 14-2, 7-3, 6-4, 11-, and A must capture on his next (12th) move, -13, 54-20, 53-27, 52-34, 51-41, 50-34, 42-27, 35-20, 28-13, 21-6, 14-2, 7-3, 6-4, 11-, and A must capture on his next (14th) move.

395.--A WAR PUZZLE GAME.

The Britisher can always catch the enemy, no matter how clever and elusive that astute individual may be; but curious though it may seem, the British general can only do so after he has paid a somewhat mysterious visit to the particular town marked "1" in the map, going in by 3 and leaving by 2, or entering by 2 and leaving by 3. The three towns that are shaded and have no numbers do not really come into the question, as some may suppose, for the simple reason that the Britisher never needs to enter any one of them, while the enemy cannot be forced to go into them, and would be clearly ill-advised to do so voluntarily. We may therefore leave these out of consideration altogether. No matter what the enemy may do, the Britisher should make the following first nine moves: He should visit towns 24, 20, 19, 15, 11, 7, 3, 1, 2. If the enemy takes it into his head also to go to town 1, it will be found that he will have to beat a precipitate retreat _the same way that he went in_, or the Britisher will infallibly catch him in towns 2 or 3, as the case may be. So the enemy will be wise to avoid that north-west corner of the map altogether.

Now, when the British general has made the nine moves that I have given, the enemy will be, after his own ninth move, in one of the towns marked 5, 8, 11, 13, 14, 16, 19, 21, 24, or 27. Of course, if he imprudently goes to 3 or 6 at this point he will be caught at once. Wherever he may happen to be, the Britisher "goes for him," and has no longer any difficulty in catching him in eight more moves at most (seventeen in all) in one of the following ways. The Britisher will get to 8 when the enemy is at 5, and win next move; or he will get to 19 when the enemy is at 22, and win next move; or he will get to 24 when the enemy is at 27, and so win next move. It will be found that he can be forced into one or other of these fatal positions.

In short, the strategy really amounts to this: the Britisher plays the first nine moves that I have given, and although the enemy does his very best to escape, our general goes after his antagonist and always driving him away from that north-west corner ultimately closes in with him, and wins. As I have said, the Britisher never need make more than seventeen moves in all, and may win in fewer moves if the enemy plays badly. But after playing those first nine moves it does not matter even if the Britisher makes a few bad ones. He may lose time, but cannot lose his advantage so long as he now keeps the enemy from town 1, and must eventually catch him.

This is a complete explanation of the puzzle. It may seem a little complex in print, but in practice the winning play will now be quite easy to the reader. Make those nine moves, and there ought to be no difficulty whatever in finding the concluding line of play. Indeed, it might almost be said that then it is difficult for the British general _not_ to catch the enemy. It is a question of what in chess we call the "opposition," and the visit by the Britisher to town 1 "gives him the jump" on the enemy, as the man in the street would say.

Here is an illustrative example in which the enemy avoids capture as long as it is possible for him to do so. The Britisher's moves are above the line and the enemy's below it. Play them alternately.

24 20 19 15 11 7 3 1 2 6 10 14 18 19 20 24
-----------------------------------------------
13 9 13 17 21 20 24 23 19 15 19 23 24 25 27

The enemy must now go to 25 or B, in either of which towns he is immediately captured.

396.--A MATCH MYSTERY.

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