The Project Gutenberg eBook of Amusements In Mathematics, by Henry Ernest Dudeney.

Total No. of Counters.	L MOVEMENT.		U MOVEMENT.		Total No. of Moves.
Total No. of Counters.	No. of Counters.	No. of Moves.	No. of Counters.	No. of Moves.	Total No. of Moves.
4n	n - 1 and n	2(n - 1)² + 5n - 7	2n + 1	2n² + 3n + 1	4(n² + n - 1)
4n - 2	n - 1 " n	2(n - 1)² + 5n - 7	2n - 1	2(n - 1)² + 3n - 2	4n² - 5
4n + 1	n " n + 1	2n² + 5n - 2	2n	2n² + 3n - 4	2(2n² + 4n - 3)
4n - 1	n - 1 " n	2(n - 1)² + 5n - 7	2n	2n² + 3n - 4	4n² + 4n - 9

SECOND METHOD.

Total No. of Counters.	L MOVEMENT.		U MOVEMENT.		Total No. of Moves.
Total No. of Counters.	No. of Counters.	No. of Moves.	No. of Counters.	No. of Moves.	Total No. of Moves.
4n	n and n	2n² + 3n - 4	2n	2(n - 1)² + 5n - 2	4(n² + n - 1)
4n - 2	n - 1 " n - 1	2(n - 1)² + 3n - 7	2n	2(n - 1)² + 5n - 2	4n² - 5
4n + 1	n " n	2n² + 3n - 4	2n + 1	2n² + 5n - 2	2(2n² + 4n - 3)
4n - 1	n " n	2n² + 3n - 4	2n - 1	2(n - 1)² + 5n - 7	4n² + 4n-9

More generally we may say that with m counters, where m is even and greater than 4, we require ^{(m² + 4m - 16)}/₄ moves; and where m is odd and greater than 3, ^{(m² + 6m - 31)}/₄ moves. I have thus shown the reader how to find the minimum number of moves for any case, and the character and direction of the moves. I will leave him to discover for himself how the actual order of moves is to be determined. This is a hard nut, and requires careful adjustment of the L and the U movements, so that they may be mutually accommodating.

The following leaps solve the puzzle in ten moves: 2 to 1, 5 to 2, 3 to 5, 6 to 3, 7 to 6, 4 to 7, 1 to 4, 3 to 1, 6 to 3, 7 to 6.

Play the counters in the following order: K C E K W T C E H M K W T A N C E H M I K C E H M T, and there you are, at Twickenham. The position itself will always determine whether you are to make a leap or a simple move.

In solving this puzzle there were two things to be achieved: first, so to manipulate the counters that the word VICTORIA should read round the cross in the same direction, only with the V on one of the dark arms; and secondly, to perform the feat in the fewest possible moves. Now, as a matter of fact, it would be impossible to perform the first part in any way whatever if all the letters of the word were different; but as there are two I's, it can be done by making these letters change places—that is, the first I changes from the 2nd place to the 7th, and the second I from the 7th place to the 2nd. But the point I referred to, when introducing the puzzle, as a little remarkable is this: that a solution in twenty-two moves is obtainable by moving the letters in the order of the following words: "A VICTOR! A VICTOR! A VICTOR I!"

There are, however, just six solutions in eighteen moves, and the following is one of them: I (1), V, A, I (2), R, O, T, I (1), I (2), A, V, I (2), I (1), C, I (2), V, A, I (1). The first and second I in the word are distinguished by the numbers 1 and 2.

It will be noticed that in the first solution given above one of the I's never moves, though the movements of the other letters cause it to change its relative position. There is another peculiarity I may point out—that there is a solution in twenty-eight moves requiring no letter to move to the central division except the I's. I may also mention that, in each of the solutions in eighteen moves, the letters C, T, O, R move once only, while the second I always moves four times, the V always being transferred to the right arm of the cross.

This puzzle can be solved in 23 moves—the fewest possible. Move the blocks in the following order: A, B, F, E, C, A, B, F, E, C, A, B, D, H, G, A, B, D, H, G, D, E, F.

The shortest possible way is to move the articles in the following order: Piano, bookcase, wardrobe, piano, cabinet, chest of drawers, piano, wardrobe, bookcase, cabinet, wardrobe, piano, chest of drawers, wardrobe, cabinet, bookcase, piano. Thus seventeen removals are necessary. The landlady could then move chest of drawers, wardrobe, and cabinet. Mr. Dobson did not mind the wardrobe and chest of drawers changing rooms so long as he secured the piano.

The solution to the Eight Engines Puzzle is as follows: The engine that has had its fire drawn and therefore cannot move is No. 5. Move the other engines in the following order: 7, 6, 3, 7, 6, 1, 2, 4, 1, 3, 8, 1, 3, 2, 4, 3, 2, seventeen moves in all, leaving the eight engines in the required order.

This little puzzle may be solved in as few as nine moves. Play the engines as follows: From 9 to 10, from 6 to 9, from 5 to 6, from 2 to 5, from 1 to 2, from 7 to 1, from 8 to 7, from 9 to 8, and from 10 to 9. You will then have engines A, B, and C on each of the three circles and on each of the three straight lines. This is the shortest solution that is possible.

Only six reversals are necessary. The white train (from A to D) is divided into three sections, engine and 7 wagons, 8 wagons, and 1 wagon. The black train (D to A) never uncouples anything throughout. Fig. 1 is original position with 8 and 1 uncoupled. The black train proceeds to position in Fig. 2 (no reversal). The engine and 7 proceed towards D, and black train backs, leaves 8 on loop, and takes up position in Fig. 3 (first reversal). Black train goes to position in Fig. 4 to fetch single wagon (second reversal). Black train pushes 8 off loop and leaves single wagon there, proceeding on its journey, as in Fig. 5 (third and fourth reversals). White train now backs on to loop to pick up single car and goes right away to D (fifth and sixth reversals).

The exchange of cars can be made in forty-three moves, as follows: 6-G, 2-B, 1-E, 3-H, 4-I, 3-L, 6-K, 4-G, 1-I, 2-J, 5-H, 4-A, 7-F, 8-E, 4-D, 8-C, 7-A, 8-G, 5-C, 2-B, 1-E, 8-I, 1-G, 2-J, 7-H, 1-A, 7-G, 2-B, 6-E, 3-H, 8-L, 3-I, 7-K, 3-G, 6-I, 2-J, 5-H, 3-C, 5-G, 2-B, 6-E, 5-I, 6-J. Of course, "6-G" means that the car numbered "6" moves to the point "G." There are other ways in forty-three moves.

It will be seen in the illustration how the prisoners may be arranged so as to produce as many as sixteen even rows. There are 4 such vertical rows, 4 horizontal rows, 5 diagonal rows in one direction, and 3 diagonal rows in the other direction. The arrows here show the movements of the four prisoners, and it will be seen that the infirm man in the bottom corner has not been moved.

In order to place words round the circle under the conditions, it is necessary to select words in which letters are repeated in certain relative positions. Thus, the word that solves our puzzle is "Swansea," in which the first and fifth letters are the same, and the third and seventh the same. We make out jumps as follows, taking the letters of the word in their proper order: 2-5, 7-2, 4-7, 1-4, 6-1, 3-6, 8-3. Or we could place a word like "Tarapur" (in which the second and fourth letters, and the third and seventh, are alike) with these moves: 6-1, 7-4, 2-7, 5-2, 8-5, 3-6, 8-3. But "Swansea" is the only word, apparently, that will fulfil the conditions of the puzzle.

This puzzle should be compared with Sharp's Puzzle, referred to in my solution to No. 341, "The Four Frogs." The condition "touch and jump over two" is identical with "touch and move along a line."

Here is a solution in nineteen moves; the moves enclosed in brackets count as one move only: 19-17, 16-18, (29-17, 17-19), 30-18, 27-25, (22-24, 24-26), 31-23, (4-16, 16-28), 7-9, 10-8, 12-10, 3-11, 18-6, (1-3, 3-11), (13-27, 27-25), (21-7, 7-9), (33-31, 31-23), (10-8, 8-22, 22-24, 24-26, 26-12, 12-10), 5-17. All the counters are now removed except one, which is left in the central hole. The solution needs judgment, as one is tempted to make several jumps in one move, where it would be the reverse of good play. For example, after playing the first 3-11 above, one is inclined to increase the length of the move by continuing with 11-25, 25-27, or with 11-9, 9-7.

Number the plates (1, 2, 3, 4), (5, 6, 7, 8), (9, 10, 11, 12), (13, 14, 15, 16) in successive rows from the top to the bottom. Then transfer the apple from 8 to 10 and play as follows, always removing the apple jumped over: 9-11, 1-9, 13-5, 16-8, 4-12, 12-10, 3-1, 1-9, 9-11.

This puzzle may be solved in as few as four moves, in the following manner: Move 5 over 8, 9, 3, 1. Move 7 over 4. Move 6 over 2 and 7. Move 5 over 6, and all the counters are removed except 5, which is left in the central square that it originally occupied.

Here is one of several solutions. Move 12 to 3, 7 to 4, 10 to 6, 8 to 1, 9 to 5, 11 to 2.

Number the plates from 1 to 12 in the order that the boy is seen to be going in the illustration. Starting from 1, proceed as follows, where "1 to 4" means that you take the coin from plate No. 1 and transfer it to plate No. 4: 1 to 4, 5 to 8, 9 to 12, 3 to 6, 7 to 10, 11 to 2, and complete the last revolution to 1, making three revolutions in all. Or you can proceed this way: 4 to 7, 8 to 11, 12 to 3, 2 to 5, 6 to 9, 10 to 1. It is easy to solve in four revolutions, but the solutions in three are more difficult to discover.

This is "The Riddle of the Fishpond" (No. 41, Canterbury Puzzles) in a different dress.

In order that the cat should eat every thirteenth mouse, and the white mouse last of all, it is necessary that the count should begin at the seventh mouse (calling the white one the first)—that is, at the one nearest the tip of the cat's tail. In this case it is not at all necessary to try starting at all the mice in turn until you come to the right one, for you can just start anywhere and note how far distant the last one eaten is from the starting point. You will find it to be the eighth, and therefore must start at the eighth, counting backwards from the white mouse. This is the one I have indicated.

In the case of the second puzzle, where you have to find the smallest number with which the cat may start at the white mouse and eat this one last of all, unless you have mastered the general solution of the problem, which is very difficult, there is no better course open to you than to try every number in succession until you come to one that works correctly. The smallest number is twenty-one. If you have to proceed by trial, you will shorten your labour a great deal by only counting out the remainders when the number is divided successively by 13, 12, 11, 10, etc. Thus, in the case of 21, we have the remainders 8, 9, 10, 1, 3, 5, 7, 3, 1, 1, 3, 1, 1. Note that I do not give the remainders of 7, 3, and 1 as nought, but as 7, 3, and 1. Now, count round each of these numbers in turn, and you will find that the white mouse is killed last of all. Of course, if we wanted simply any number, not the smallest, the solution is very easy, for we merely take the least common multiple of 13, 12, 11, 10, etc. down to 2. This is 360360, and you will find that the first count kills the thirteenth mouse, the next the twelfth, the next the eleventh, and so on down to the first. But the most arithmetically inclined cat could not be expected to take such a big number when a small one like twenty-one would equally serve its purpose.

In the third case, the smallest number is 100. The number 1,000 would also do, and there are just seventy-two other numbers between these that the cat might employ with equal success.

To leave the three piles at the extreme ends of the rows, the cheeses may be moved as follows—the numbers refer to the cheeses and not to their positions in the row: 7-2, 8-7, 9-8, 10-15, 6-10, 5-6, 14-16, 13-14, 12-13, 3-1, 4-3, 11-4. This is probably the easiest solution of all to find. To get three of the piles on cheeses 13, 14, and 15, play thus: 9-4, 10-9, 11-10, 6-14, 5-6, 12-15, 8-12, 7-8, 16-5, 3-13, 2-3, 1-2. To leave the piles on cheeses 3, 5, 12, and 14, play thus: 8-3, 9-14, 16-12, 1-5, 10-9, 7-10, 11-8, 2-1, 4-16, 13-2, 6-11, 15-4.

Make the following exchanges of pairs: H-K, H-E, H-C, H-A, I-L, I-F, I-D, K-L, G-J, J-A, F-K, L-E, D-K, E-F, E-D, E-B, B-K. It will be found that, although the white counters can be moved to their proper places in 11 moves, if we omit all consideration of exchanges, yet the black cannot be so moved in fewer than 17 moves. So we have to introduce waste moves with the white counters to equal the minimum required by the black. Thus fewer than 17 moves must be impossible. Some of the moves are, of course, interchangeable.

If the enemy's fleet be anchored in the formation shown in the illustration, it will be seen that as many as ten out of the sixteen ships may be blown up by discharging the torpedoes in the order indicated by the numbers and in the directions indicated by the arrows. As each torpedo in succession passes under three ships and sinks the fourth, strike out each vessel with the pencil as it is sunk.

I suggested that the reader should try this puzzle with counters, so I give my solution in that form. The silk hats are represented by black counters and the felt hats by white counters. The first row shows the hats in their original positions, and then each successive row shows how they appear after one of the five manipulations. It will thus be seen that we first move hats 2 and 3, then 7 and 8, then 4 and 5, then 10 and 11, and, finally, 1 and 2, leaving the four silk hats together, the four felt hats together, and the two vacant pegs at one end of the row. The first three pairs moved are dissimilar hats, the last two pairs being similar. There are other ways of solving the puzzle.

There are a good many different solutions to this puzzle. Any contiguous pair, except 7-8, may be moved first, and after the first move there are variations. The following solution shows the position from the start right through each successive move to the end:—

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

Two of the pots, 13 and 19, were in their proper places. As every interchange may result in a pot being put in its place, it is clear that twenty-two interchanges will get them all in order. But this number of moves is not the fewest possible, the correct answer being seventeen. Exchange the following pairs: (3-1, 2-3), (15-4, 16-15), (17-7, 20-17), (24-10, 11-24, 12-11), (8-5, 6-8, 21-6, 23-21, 22-23, 14-22, 9-14, 18-9). When you have made the interchanges within any pair of brackets, all numbers within those brackets are in their places. There are five pairs of brackets, and 5 from 22 gives the number of changes required—17.

As the conditions are generally understood, this puzzle is incapable of solution. This can be demonstrated quite easily. So we have to look for some catch or quibble in the statement of what we are asked to do. Now if you fold the paper and then push the point of your pencil down between the fold, you can with one stroke make the two lines CD and EF in our diagram. Then start at A, and describe the line ending at B. Finally put in the last line GH, and the thing is done strictly within the conditions, since folding the paper is not actually forbidden. Of course the lines are here left unjoined for the purpose of clearness.

In the rubbing out form of the puzzle, first rub out A to B with a single finger in one stroke. Then rub out the line GH with one finger. Finally, rub out the remaining two vertical lines with two fingers at once! That is the old trick.

There are just sixteen points (all on the outside) where three roads may be said to join. These are called by mathematicians "odd nodes." There is a rule that tells us that in the case of a drawing like the present one, where there are sixteen odd nodes, it requires eight separate strokes or routes (that is, half as many as there are odd nodes) to complete it. As we have to produce as much as possible with only one of these eight strokes, it is clearly necessary to contrive that the seven strokes from odd node to odd node shall be as short as possible. Start at A and end at B, or go the reverse way.

It can be done in twelve continuous strokes, thus: Start at A in the illustration, and eight strokes, forming the star, will bring you back to A; then one stroke round the circle to B, one stroke to C, one round the circle to D, and one final stroke to E—twelve in all. Of course, in practice the second circular stroke will be over the first one; it is separated in the diagram, and the points of the star not joined to the circle, to make the solution clear to the eye.

The inspector need only travel nineteen miles if he starts at B and takes the following route: B A D G D E F I F C B E H K L I H G J K. Thus the only portions of line travelled over twice are the two sections D to G and F to I. Of course, the route may be varied, but it cannot be shortened.

Note that there are six towns, from which only two roads issue. Thus 1 must lie between 9 and 12 in the circular route. Mark these two roads as settled. Similarly mark 9, 5, 14, and 4, 8, 14, and 10, 6, 15, and 10, 2, 13, and 3, 7, 13. All these roads must be taken. Then you will find that he must go from 4 to 15, as 13 is closed, and that he is compelled to take 3, 11, 16, and also 16, 12. Thus, there is only one route, as follows: 1, 9, 5, 14, 8, 4, 15, 6, 10, 2, 13, 7, 3, 11, 16, 12, 1, or its reverse—reading the line the other way. Seven roads are not used.

It will be seen from the illustration (where the roads not used are omitted) that the traveller can go as far as seventy miles in fifteen turnings. The turnings are all numbered in the order in which they are taken. It will be seen that he never visits nineteen of the towns. He might visit them all in fifteen turnings, never entering any town twice, and end at the black town from which he starts (see "The Rook's Tour," No. 320), but such a tour would only take him sixty-four miles.

Though we cannot really see all the sides of the octahedron at once, we can make a projection of it that suits our purpose just as well. In the diagram the six points represent the six angles of the octahedron, and four lines proceed from every point under exactly the same conditions as the twelve edges of the solid. Therefore if we start at the point A and go over all the lines once, we must always end our route at A. And the number of different routes is just 1,488, counting the reverse way of any route as different. It would take too much space to show how I make the count. It can be done in about five minutes, but an explanation of the method is difficult. The reader is therefore asked to accept my answer as correct.

There are thirty edges, of which eighteen were visible in the original illustration, represented in the following diagram by the hexagon NAESGD. By this projection of the solid we get an imaginary view of the remaining twelve edges, and are able to see at once their direction and the twelve points at which all the edges meet. The difference in the length of the lines is of no importance; all we want is to present their direction in a graphic manner. But in case the novice should be puzzled at only finding nineteen triangles instead of the required twenty, I will point out that the apparently missing triangle is the outline HIK.

In this case there are twelve odd nodes; therefore six distinct and disconnected routes will be needful if we are not to go over any lines twice. Let us therefore find the greatest distance that we may so travel in one route.

It will be noticed that I have struck out with little cross strokes five lines or edges in the diagram. These five lines may be struck out anywhere so long as they do not join one another, and so long as one of them does not connect with N, the North Pole, from which we are to start. It will be seen that the result of striking out these five lines is that all the nodes are now even except N and S. Consequently if we begin at N and stop at S we may go over all the lines, except the five crossed out, without traversing any line twice. There are many ways of doing this. Here is one route: N to H, I, K, S, I, E, S, G, K, D, H, A, N, B, A, E, F, B, C, G, D, N, C, F, S. By thus making five of the routes as short as is possible—simply from one node to the next—we are able to get the greatest possible length for our sixth line. A greater distance in one route, without going over the same ground twice, it is not possible to get.

It is now readily seen that those five erased lines must be gone over twice, and they may be "picked up," so to speak, at any points of our route. Thus, whenever the traveller happens to be at I he can run up to A and back before proceeding on his route, or he may wait until he is at A and then run down to I and back to A. And so with the other lines that have to be traced twice. It is, therefore, clear that he can go over 25 of the lines once only (25 × 10,000 miles = 250,000 miles) and 5 of the lines twice (5 × 20,000 miles = 100,000 miles), the total, 350,000 miles, being the length of his travels and the shortest distance that is possible in visiting the whole body.

It will be noticed that I have made him end his travels at S, the South Pole, but this is not imperative. I might have made him finish at any of the other nodes, except the one from which he started. Suppose it had been required to bring him home again to N at the end of his travels. Then instead of suppressing the line AI we might leave that open and close IS. This would enable him to complete his 350,000 miles tour at A, and another 10,000 miles would take him to his own fireside. There are a great many different routes, but as the lengths of the edges are all alike, one course is as good as another. To make the complete 350,000 miles tour from N to S absolutely clear to everybody, I will give it entire: N to H, I, A, I, K, H, K, S, I, E, S, G, F, G, K, D, C, D, H, A, N, B, E, B, A, E, F, B, C, G, D, N, C, F, S—that is, thirty-five lines of 10,000 miles each.

Starting from A, the inspector need only travel 36 furlongs if he takes the following route: A to B, G, H, C, D, I, H, M, N, I, J, O, N, S, R, M, L, G, F, K, L, Q, R, S, T, O, J, E, D, C, B, A, F, K, P, Q. He thus passes between A and B twice, between C and D twice, between F and K twice, between J and O twice, and between R and S twice—five repetitions. Therefore 31 passages plus 5 repeated equal 36 furlongs. The little pitfall in this puzzle lies in the fact that we start from an even node. Otherwise we need only travel 35 furlongs.

When Mr. Maggs replied, "No way, I'm sure," he was not saying that the thing was impossible, but was really giving the actual route by which the problem can be solved. Starting from the star, if you visit the towns in the order, NO WAY, I'M SURE, you will visit every town once, and only once, and end at E. So both men were correct. This was the little joke of the puzzle, which is not by any means difficult.

There are only four different routes (or eight, if we count the reverse ways) by which the sailor can start at the island marked A, visit all the islands once, and once only, and return again to A. Here they are:—

A I P T L O E H R Q D C F U G N S K M B A A I P T S N G L O E U F C D K M B Q R H A A B M K S N G L T P I O E U F C D Q R H A A I P T L O E U G N S K M B Q D C F R H A

Now, if the sailor takes the first route he will make C his 12th island (counting A as 1); by the second route he will make C his 13th island; by the third route, his 16th island; and by the fourth route, his 17th island. If he goes the reverse way, C will be respectively his 10th, 9th, 6th, and 5th island. As these are the only possible routes, it is evident that if the sailor puts off his visit to C as long as possible, he must take the last route reading from left to right. This route I show by the dark lines in the diagram, and it is the correct answer to the puzzle.

The map may be greatly simplified by the "buttons and string" method, explained in the solution to No. 341, "The Four Frogs."

The first thing to do in trying to solve a puzzle like this is to attempt to simplify it. If you look at Fig. 1, you will see that it is a simplified version of the map. Imagine the circular towns to be buttons and the railways to be connecting strings. (See solution to No. 341.) Then, it will be seen, we have simply "straightened out" the previous diagram without affecting the conditions. Now we can further simplify by converting Fig. 1 into Fig. 2, which is a portion of a chessboard. Here the directions of the railways will resemble the moves of a rook in chess—that is, we may move in any direction parallel to the sides of the diagram, but not diagonally. Therefore the first town (or square) visited must be a black one; the second must be a white; the third must be a black; and so on. Every odd square visited will thus be black and every even one white. Now, we have 23 squares to visit (an odd number), so the last square visited must be black. But Z happens to be white, so the puzzle would seem to be impossible of solution.

As we were told that the man "succeeded" in carrying put his plan, we must try to find some loophole in the conditions. He was to "enter every town once and only once," and we find no prohibition against his entering once the town A after leaving it, especially as he has never left it since he was born, and would thus be "entering" it for the first time in his life. But he must return at once from the first town he visits, and then he will have only 22 towns to visit, and as 22 is an even number, there is no reason why he should not end on the white square Z. A possible route for him is indicated by the dotted line from A to Z. This route is repeated by the dark lines in Fig. 1, and the reader will now have no difficulty in applying; it to the original map. We have thus proved that the puzzle can only be solved by a return to A immediately after leaving it.

According to the conditions, in the strict sense in which one at first understands them, thereis no possible solution to this puzzle. In such a dilemma one always has to look for some verbal quibble or trick. If the owner of house A will allow the water company to run their pipe for house C through his property (and we are not bound to assume that he would object), then the difficulty is got over, as shown in our illustration. It will be seen that the dotted line from W to C passes through house A, but no pipe ever crosses another pipe.

The routes taken by the eight drivers are shown in the illustration, where the dotted line roads are omitted to make the paths clearer to the eye.

The simplest way is to write in the number of routes to all the towns in this manner. Put a 1 on all the towns in the top row and in the first column. Then the number of routes to any town will be the sum of the routes to the town immediately above and to the town immediately to the left. Thus the routes in the second row will be 1, 2, 3, 4, 5, 6, etc., in the third row, 1, 3, 6, 10, 15, 21, etc.; and so on with the other rows. It will then be seen that the only town to which there are exactly 1,365 different routes is the twelfth town in the fifth row—the one immediately over the letter E. This town was therefore the cyclist's destination.

The general formula for the number of routes from one corner to the corner diagonally opposite on any such rectangular reticulated arrangement, under the conditions as to direction, is ^(m + n)!/_m!n!, where m is the number of towns on one side, less one, and n the number on the other side, less one. Our solution involves the case where there are 12 towns by 5. Therefore m = 11 and n = 4. Then the formula gives us the answer 1,365 as above.

First of all I will ask the reader to compare the original square diagram with the circular one shown in Figs. 1, 2, and 3 below. If for the moment we ignore the shading (the purpose of which I shall proceed to explain), we find that the circular diagram in each case is merely a simplification of the original square one—that is, the roads from A lead to B, E, and M in both cases, the roads from L (London) lead to I, K, and S, and so on. The form below, being circular and symmetrical, answers my purpose better in applying a mechanical solution, and I therefore adopt it without altering in any way the conditions of the puzzle. If such a question as distances from town to town came into the problem, the new diagrams might require the addition of numbers to indicate these distances, or they might conceivably not be at all practicable.

Now, I draw the three circular diagrams, as shown, on a sheet of paper and then cut out three pieces of cardboard of the forms indicated by the shaded parts of these diagrams. It can be shown that every route, if marked out with a red pencil, will form one or other of the designs indicated by the edges of the cards, or a reflection thereof. Let us direct our attention to Fig. 1. Here the card is so placed that the star is at the town T; it therefore gives us (by following the edge of the card) one of the circular routes from London: L, S, R, T, M, A, E, P, O, J, D, C, B, G, N, Q, K, H, F, I, L. If we went the other way, we should get L, I, F, H, K, Q, etc., but these reverse routes were not to be counted. When we have written out this first route we revolve the card until the star is at M, when we get another different route, at A a third route, at E a fourth route, and at P a fifth route. We have thus obtained five different routes by revolving the card as it lies. But it is evident that if we now take up the card and replace it with the other side uppermost, we shall in the same manner get five other routes by revolution.

We therefore see how, by using the revolving card in Fig. 1, we may, without any difficulty, at once write out ten routes. And if we employ the cards in Figs. 2 and 3, we similarly obtain in each case ten other routes. These thirty routes are all that are possible. I do not give the actual proof that the three cards exhaust all the possible cases, but leave the reader to reason that out for himself. If he works out any route at haphazard, he will certainly find that it falls into one or other of the three categories.

Let us confine our attention to the L in the top left-hand corner. Suppose we go by way of the E on the right: we must then go straight on to the V, from which letter the word may be completed in four ways, for there are four E's available through which we may reach an L. There are therefore four ways of reading through the right-hand E. It is also clear that there must be the same number of ways through the E that is immediately below our starting point. That makes eight. If, however, we take the third route through the E on the diagonal, we then have the option of any one of the three V's, by means of each of which we may complete the word in four ways. We can therefore spell LEVEL in twelve ways through the diagonal E. Twelve added to eight gives twenty readings, all emanating from the L in the top left-hand corner; and as the four corners are equal, the answer must be four times twenty, or eighty different ways.

There are 252 different ways. The general formula is that, for words of n letters (not palindromes, as in the case of the next puzzle), when grouped in this manner, there are always 2⁽ⁿ⁺¹⁾ - 4 different readings. This does not allow diagonal readings, such as you would get if you used instead such a word as DIGGING, where it would be possible to pass from one G to another G by a diagonal step.

The correct answer is 1,992 different ways. Every F is either a corner F or a side F—standing next to a corner in its own square of F's. Now, FIED may be read from a corner F in 16 ways; therefore DEIF may be read into a corner F also in 16 ways; hence DEIFIED may be read through a corner F in 16 × 16 = 256 ways. Consequently, the four corner F's give 4 × 256 = 1,024 ways. Then FIED may be read from a side F in 11 ways, and DEIFIED therefore in 121 ways. But there are eight side F's; consequently these give together 8 × 121 = 968 ways. Add 968 to 1,024 and we get the answer, 1,992.

In this form the solution will depend on whether the number of letters in the palindrome be odd or even. For example, if you apply the word NUN in precisely the same manner, you will get 64 different readings; but if you use the word NOON, you will only get 56, because you cannot use the same letter twice in immediate succession (since you must "always pass from one letter to another") or diagonal readings, and every reading must involve the use of the central N.

The reader may like to find for himself the general formula in this case, which is complex and difficult. I will merely add that for such a case as MADAM, dealt with in the same way as DEIFIED, the number of readings is 400.

THE number of readings here is 63,504, as in the case of "WAS IT A RAT I SAW" (No. 30, Canterbury Puzzles). The general formula is that for palindromic sentences containing 2n + 1 letters there are [4(2ⁿ - 1)]² readings.

Starting from any one of the N's, there are 17 different readings of NAH, or 68 (4 times 17) for the 4 N's. Therefore there are also 68 ways of spelling HAN. If we were allowed to use the same N twice in a spelling, the answer would be 68 times 68, or 4,624 ways. But the conditions were, "always passing from one letter to another." Therefore, for every one of the 17 ways of spelling HAN with a particular N, there would be 51 ways (3 times 17) of completing the NAH, or 867 (17 times 51) ways for the complete word. Hence, as there are four N's to use in HAN, the correct solution of the puzzle is 3,468 (4 times 867) different ways.

The required proverb is, "There is many a slip 'twixt the cup and the lip." Start at the T on the outside at the bottom right-hand corner, pass to the H above it, and the rest is easy.

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

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

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