Chapter 36

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.

+-+-+-+-+-+-+-+-+ |r|n| | |k| |n|r| +-+-+-+-+-+-+-+-+ | |p| | | | |p|p| +-+-+-+-+-+-+-+-+ | | | |p| | | | | +-+-+-+-+-+-+-+-+ |p| |p|P| | | | | +-+-+-+-+-+-+-+-+ |P|b|P| | |p| |q| +-+-+-+-+-+-+-+-+ | |b| | |p|P|R|P| +-+-+-+-+-+-+-+-+ | |P| |N|P| |P|Q| +-+-+-+-+-+-+-+-+ | | |B| |K|B|N|R| +-+-+-+-+-+-+-+-+

350.--THE FORSAKEN KING.

Play as follows:--

White. Black. 1. P to K 4th 1. Any move 2. Q to Kt 4th 2. Any move except on KB file (a) 3. Q to Kt 7th 3. K moves to royal row 4. B to Kt 5th 4. Any move 5. Mate in two moves If 3. K other than to royal row 4. P to Q 4th 4. Any move 5. Mate in two moves (a) If 2. Any move on KB file 3. Q to Q 7th 3. K moves to royal row 4. P to Q Kt 3rd 4. Any move 5. Mate in two moves If 3. K other than to royal row 4. P to Q 4th 4. Any move 5. Mate in two moves

Of course, by "royal row" is meant the row on which the king originally stands at the beginning of a game. Though, if Black plays badly, he may, in certain positions, be mated in fewer moves, the above provides for every variation he can possibly bring about.

351.--THE CRUSADER.

White. Black. 1. Kt to QB 3rd 1. P to Q 4th 2. Kt takes QP 2. Kt to QB 3rd 3. Kt takes KP 3. P to KKt 4th 4. Kt takes B 4. Kt to KB 3rd 5. Kt takes P 5. Kt to K 5th 6. Kt takes Kt 6. Kt to B 6th 7. Kt takes Q 7. R to KKt sq 8. Kt takes BP 8. R to KKt 3rd 9. Kt takes P 9. R to K 3rd 10. Kt takes P 10. Kt to Kt 8th 11. Kt takes B 11. R to R 6th 12. Kt takes R 12. P to Kt 4th 13. Kt takes P (ch) 13. K to B 2nd 14. Kt takes P 14. K to Kt 3rd 15. Kt takes R 15. K to R 4th 16. Kt takes Kt 16. K to R 5th White now mates in three moves. 17. P to Q 4th 17. K to R 4th 18. Q to Q 3rd 18. K moves 19. Q to KR 3rd (mate) If 17. K to Kt 5th 18. P to K 4th (dis. ch) 18. K moves 19. P to KKt 3rd (mate)

The position after the sixteenth move, with the mate in three moves, was first given by S. Loyd in _Chess Nuts_.

352.--IMMOVABLE PAWNS.

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.

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

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.