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).
224.—THE MOTOR-GARAGE PUZZLE.
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—1, 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.
225.—THE TEN PRISONERS.
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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 comer has not been moved.
226.—ROUND THE COAST.
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."
227.—CENTRAL SOLITAIRE.
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.
I do not think the number of moves can be reduced.
228.—THE TEN APPLES.
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.
229.—THE NINE ALMONDS.
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.
230.—THE TWELVE PENNIES.
Here is one of several solutions. Move 12 to 3, 7 to 4, 10 to 6, 8 to I, 9 to 5, 11 to 2.
231.—PLATES AND COINS.
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.
