| 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.
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 circumstances, send his wife over to the other bank to a man or men, even if she assured 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 passages from land to land (though French mathematicians have declared in their books that in such circum- stances 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 en- gaged. The three columns show the positions of the different individuals on the lawn, the island, and the opposite shore before starting and after each passage, while the asterisk indi- cates the position of the boat on every occasion.
Having found the fewest possible passages, 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,
| Lawn. | Island. | Shore. | |||||
|---|---|---|---|---|---|---|---|
| ABCD | abcd | * | nbsp; | ||||
| ABCD | cd | ab | * | ||||
| ABCD | bcd | * | a | ||||
| ABCD | d | bc | * | a | |||
and in the solution I have given (where the girls do 8-i3ths 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 ex- changes 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
