succes- 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.
237.—BOYS AND GIRLS.
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 | 7 | . | . | 5 | 6 | 8 |
| 4 | 3 | 1 | 2 | . | . | 5 | 6 | 7 | 8 | 4 | . | . | 2 | 7 | 1 | 3 | 5 | 6 | 8 |
| 4 | 3 | 1 | 2 | 7 | 6 | 5 | . | . | 8 | 4 | 8 | 6 | 2 | 7 | 1 | 3 | 5 | . | . |
238.—ARRANGING THE JAM POTS.
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.
239.—A JUVENILE PUZZLE.
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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 C D and E F 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.
240.—THE UNION JACK.
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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.
241.—THE DISSECTED CIRCLE.
It can be done in twelve continuous strokes, thus: Start at A in the illustration, and eight
