(related to Problem: 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 circumstances 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 shore before starting and after each passage, while the asterisk indicates the position of the boat on every occasion.
| Lawn | Island | Shore |
|---|---|---|
| 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 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, and in the solution, I have given (where the girls do $8-13$ths 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.
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