Amusements in Mathematics Part 30

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 s.p.a.ce 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 s.p.a.ce 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 3 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 quant.i.ty 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 a.s.sumes 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 circ.u.mstances, send his wife over to the other bank to a man or men, even if she a.s.sured 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 pa.s.sages from land to land (though French mathematicians have declared in their books that in such circ.u.mstances 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 sh.o.r.e before starting and after each pa.s.sage, while the asterisk indicates the position of the boat on every occasion.

Lawn. Island. Sh.o.r.e. ABCDabcd * ABCD cd ab * ABCD bcd * a ABCD d bc * a ABCD cd * b a CD cd b AB a * BCD cd * b A a BCD bcd * A a BCD d * bc A a D d bc ABC a * D d abc * ABC D d b ABC a c * B D d * b A C a c d b ABCD a c * d bc * ABCD a d ABCD abc * cd * ABCD ab ABCD abcd *

Having found the fewest possible pa.s.sages, 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.

[Ill.u.s.tration: +---+-------+-------+-------+-------+-------+-------+-------+ 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 ill.u.s.tration 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:-- [Ill.u.s.tration: +-------+-------+-------+ 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 s.p.a.ce, 2 S on s.p.a.ce, 3 D on s.p.a.ce, 2 S on 3 D, 1 H on 2 S, 2 C on s.p.a.ce, 1 D on 2 C, 4 S on s.p.a.ce, 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 s.p.a.ce (3), 4 C up on 5 H (3), 2 C up on 3 D (3), 7 D on s.p.a.ce (1), 6 C up on 7 D (3), 8 S on s.p.a.ce (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.

[Ill.u.s.tration: 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. a.s.suming 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.

[Ill.u.s.tration]

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 ill.u.s.trative 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.

If you form the three heaps (and are therefore the second to draw), any one of the following thirteen groupings will give you a win if you play correctly: 15, 14, 1; 15, 13, 2; 15, 12, 3; 15, 11, 4; 15, 10, 5; 15, 9, 6; 15, 8, 7; 14, 13, 3; 14, 11, 5; 14, 9, 7; 13, 11, 6; 13, 10, 7; 12, 11, 7.

The beautiful general solution of this problem is as follows. Express the number in every heap in powers of 2, avoiding repet.i.tions and remembering that 2^0 = 1. Then if you so leave the matches to your opponent that there is an even number of every power, you can win. And if at the start you leave the powers even, you can always continue to do so throughout the game. Take, as example, the last grouping given above--12, 11, 7. Expressed in powers of 2 we have-- 12 = 8 4 - - 11 = 8 - 2 1 7 = - 4 2 1 ------- 2 2 2 2 ------- As there are thus two of every power, you must win. Say your opponent takes 7 from the 12 heap. He then leaves-- 5 = - 4 - 1 11 = 8 - 2 1 7 = - 4 2 1 ------- 1 2 2 3 ------- Here the powers are not all even in number, but by taking 9 from the 11 heap you immediately restore your winning position, thus-- 5 = - 4 - 1 2 = - - 2 - 7 = - 4 2 1 ------- - 2 2 2 ------- And so on to the end. This solution is quite general, and applies to any number of matches and any number of heaps. A correspondent informs me that this puzzle game was first propounded by Mr. W.M.F. Mellor, but when or where it was published I have not been able to ascertain.

397.--THE MONTENEGRIN DICE GAME.

The players should select the pairs 5 and 9, and 13 and 15, if the chances of winning are to be quite equal. There are 216 different ways in which the three dice may fall. They may add up 5 in 6 different ways and 9 in 25 different ways, making 31 chances out of 216 for the player who selects these numbers. Also the dice may add up 13 in 21 different ways, and 15 in 10 different ways, thus giving the other player also 31 chances in 216.

398.--THE CIGAR PUZZLE.

Not a single member of the club mastered this puzzle, and yet I shall show that it is so simple that the merest child can understand its solution--when it is pointed out to him! The large majority of my friends expressed their entire bewilderment. Many considered that "the theoretical result, in any case, is determined by the relationship between the table and the cigars;" others, regarding it as a problem in the theory of Probabilities, arrived at the conclusion that the chances are slightly in favour of the first or second player, as the case may be. One man took a table and a cigar of particular dimensions, divided the table into equal sections, and proceeded to make the two players fill up these sections so that the second player should win. But why should the first player be so accommodating? At any stage he has only to throw down a cigar obliquely across several of these sections entirely to upset Mr. 2's calculations! We have to a.s.sume that each player plays the best possible; not that one accommodates the other.

The theories of some other friends would be quite sound if the shape of the cigar were that of a torpedo--perfectly symmetrical and pointed at both ends.

I will show that the first player should infallibly win, if he always plays in the best possible manner. Examine carefully the following diagram, No. 1, and all will be clear.

[Ill.u.s.tration: 1]

[Ill.u.s.tration: 2]

The first player must place his first cigar on end in the exact centre of the table, as indicated by the little circle. Now, whatever the second player may do throughout, the first player must always repeat it in an exactly diametrically opposite position. Thus, if the second player places a cigar at A, I put one at AA; he places one at B, I put one at BB; he places one at C, I put one at CC; he places one at D, I put one at DD; he places one at E, I put one at EE; and so on until no more cigars can be placed without touching. As the cigars are supposed to be exactly alike in every respect, it is perfectly clear that for every move that the second player may choose to make, it is possible exactly to repeat it on a line drawn through the centre of the table. The second player can always duplicate the first player's move, no matter where he may place a cigar, or whether he places it on end or on its side. As the cigars are all alike in every respect, one will obviously balance over the edge of the table at precisely the same point as another. Of course, as each player is supposed to play in the best possible manner, it becomes a matter of theory. It is no valid objection to say that in actual practice one would not be sufficiently exact to be sure of winning. If as the first player you did not win, it would be in consequence of your not having played the best possible.

The second diagram will serve to show why the first cigar must be placed on end. (And here I will say that the first cigar that I selected from a box I was able so to stand on end, and I am allowed to a.s.sume that all the other cigars would do the same.) If the first cigar were placed on its side, as at F, then the second player could place a cigar as at G--as near as possible, but not actually touching F. Now, in this position you cannot repeat his play on the opposite side, because the two ends of the cigar are not alike. It will be seen that GG, when placed on the opposite side in the same relation to the centre, intersects, or lies on top of, F, whereas the cigars are not allowed to touch. You must therefore put the cigar farther away from the centre, which would result in your having insufficient room between the centre and the bottom left-hand corner to repeat everything that the other player would do between G and the top right-hand corner. Therefore the result would not be a certain win for the first player.

399.--THE TROUBLESOME EIGHT.

[Ill.u.s.tration: +---+---+---+ 4 8 2 +---+---+---+ 3 5 7 +---+---+---+ 7 2 5 +---+---+---+ ]

The conditions were to place a different number in each of the nine cells so that the three rows, three columns, and two diagonals should each add up 15. Probably the reader at first set himself an impossible task through reading into these conditions something which is not there--a common error in puzzle-solving. If I had said "a different figure," instead of "a different number," it would have been quite impossible with the 8 placed anywhere but in a corner. And it would have been equally impossible if I had said "a different whole number." But a number may, of course, be fractional, and therein lies the secret of the puzzle. The arrangement shown in the figure will be found to comply exactly with the conditions: all the numbers are different, and the square adds up 15 in all the required eight ways.

400.--THE MAGIC STRIPS.

There are of course six different places between the seven figures in which a cut may be made, and the secret lies in keeping one strip intact and cutting each of the other six in a different place. After the cuts have been made there are a large number of ways in which the thirteen pieces may be placed together so as to form a magic square. Here is one of them:-- [Ill.u.s.tration: +-------------+ 1 2 3 4 5 6 7 +---------+---+ 3 4 5 6 7 1 2 +-----+---+---+ 5 6 7 1 2 3 4 +-+---+-------+ 7 1 2 3 4 5 6 +-+---------+-+ 2 3 4 5 6 7 1 +-------+---+-+ 4 5 6 7 1 2 3 +---+---+-----+ 6 7 1 2 3 4 5 +---+---------+ ]

The arrangement has some rather interesting features. It will be seen that the uncut strip is at the top, but it will be found that if the bottom row of figures be placed at the top the numbers will still form a magic square, and that every successive removal from the bottom to the top (carrying the uncut strip stage by stage to the bottom) will produce the same result. If we imagine the numbers to be on seven complete perpendicular strips, it will be found that these columns could also be moved in succession from left to right or from right to left, each time producing a magic square.

401.--EIGHT JOLLY GAOL-BIRDS.

There are eight ways of forming the magic square--all merely different aspects of one fundamental arrangement. Thus, if you give our first square a quarter turn you will get the second square; and as the four sides may be in turn brought to the top, there are four aspects. These four in turn reflected in a mirror produce the remaining four aspects. Now, of these eight arrangements only four can possibly be reached under the conditions, and only two of these four can be reached in the fewest possible moves, which is nineteen. These two arrangements are shown. Move the men in the following order: 5, 3, 2, 5, 7, 6, 4, 1, 5, 7, 6, 4, 1, 6, 4, 8, 3, 2, 7, and you get the first square. Move them thus: 4, 1, 2, 4, 1, 6, 7, 1, 5, 8, 1, 5, 6, 7, 5, 6, 4, 2, 7, and you have the arrangement in the second square. In the first case every man has moved, but in the second case the man numbered 3 has never left his cell. Therefore No. 3 must be the obstinate prisoner, and the second square must be the required arrangement.

[Ill.u.s.tration: +---+---+---+ +---+---+---+ 5 7 7 4 3 +- -+- -+- -+ +- -+- -+- -+ 6 4 2 4 8 +- -+- -+- -+ +- -+- -+- -+ 1 8 3 5 6 1 +---+---+---+ +---+---+---+ ]

402.--NINE JOLLY GAOL BIRDS.

There is a pitfall set for the unwary in this little puzzle. At the start one man is allowed to be placed on the shoulders of another, so as to give always one empty cell to enable the prisoners to move about without any two ever being in a cell together. The two united prisoners are allowed to add their numbers together, and are, of course, permitted to remain together at the completion of the magic square. But they are obviously not compelled so to remain together, provided that one of the pair on his final move does not break the condition of entering a cell already occupied. After the acute solver has noticed this point, it is for him to determine which method is the better one--for the two to be together at the count or to separate. As a matter of fact, the puzzle can be solved in seventeen moves if the men are to remain together; but if they separate at the end, they may actually save a move and perform the feat in sixteen! The trick consists in placing the man in the centre on the back of one of the corner men, and then working the pair into the centre before their final separation.

[Ill.u.s.tration: A B +---+---+---+ +---+---+---+ 2 9 4 6 7 2 +- -+- -+- -+ +- -+- -+- -+ 7 5 3 1 5 9 +- -+- -+- -+ +- -+- -+- -+ 6 1 8 8 3 4 +---+---+---+ +---+---+---+ ]

Here are the moves for getting the men into one or other of the above two positions. The numbers are those of the men in the order in which they move into the cell that is for the time being vacant. The pair is shown in brackets:-- Place 5 on 1. Then, 6, 9, 8, 6, 4, (6), 2, 4, 9, 3, 4, 9, (6), 7, 6, 1.

Place 5 on 9. Then, 4, 1, 2, 4, 6, (14), 8, 6, 1, 7, 6, 1, (14), 3, 4, 9.