Of course all good puzzlists are familiar with the time-honored problem of the man who had to ferry a fox, a goose, and some corn across a river in a boat which would carry but two at a time. The story of the four elopements, equally old, is built upon similar lines, but presents so many complications that the best or shortest answer seems to have been overlooked by mathematicians who have considered the subject.
It is told that four men eloped with their sweethearts, but in carrying out their plan were compelled to cross a stream in a boat which would hold but two persons at a time. In the middle of the stream, as shown in the sketch, there is a small island. It appears that the young men were so extremely jealous that not one of them would permit his prospective bride to remain at any time in the company of any other man or men unless he was also present.
Nor was any man to get into a boat alone when there happened to be a girl alone, on the island or shore, other than the one to whom he was engaged. This leads one to suspect that the girls were also jealous and feared that their fellows would run off with the wrong girl if they got a chance. Well, be that as it may, the problem is to guess the quickest way to get the whole party across the river.
Let us suppose the river to be two hundred yards wide, with an island in the middle on which any number can stand. How many trips would the boat make to get the four couples safely across in accordance with the imposed conditions?
Solution:
The feat can be performed in 17 trips.
We begin with ABCD (the men) and abcd (the girls) all on shore. The following chart is self-explanatory.
Shore Island Other side
1 A B C D c d a b
2 A B C D b c d a
3 A B C D d b c a
4 A B C D c d b a
5 C D c d b A B a
6 B C D c d b A a
7 B C D b c d A a
8 B C D d b c A a
9 D d b c A B C a
10 D d a b c A B C
11 D d b A B C a c
12 B D d b A C a c
13 d b A B C D a c
14 d b c A B C D a
15 d A B C D a b c
16 c d A B C D a b
17 A B C D a b c d
Note: There are other ways the puzzle can be solved in 17 trips. This solution is the most satisfactory in that it involves the fewest "gettings in" and "gettings out".