Solution
The earliest solution known to the jealous husbands problem, using 11 one-way trips, is as follows. The married couples are represented as (male) and a (female), and b, and and c., p. 291.
| Trip number | Starting bank | Travel | Ending bank |
|---|---|---|---|
| (start) | a b c | ||
| 1 | b c | a → | |
| 2 | b c | ← | a |
| 3 | bc → | a | |
| 4 | ← a | b c | |
| 5 | a | → | b c |
| 6 | a | ← b | c |
| 7 | a b | → | c |
| 8 | a b | ← c | |
| 9 | b | a c → | |
| 10 | b | ← | a c |
| 11 | b → | a c | |
| (finish) | a b c |
This is a shortest solution to the problem, but is not the only shortest solution., p. 291.
If however, only one man can get out of the boat at a time and husbands must be on the shore to count as with his wife as opposed to just being in the boat at the shore: move 5 to 6 is impossible, for as soon as has stepped out b on the shore won't be with her husband, despite him being just in the boat.
As mentioned previously, this solution to the jealous husbands problem will become a solution to the missionaries and cannibals problem upon replacing men by missionaries and women by cannibals. In this case we may neglect the individual identities of the missionaries and cannibals. The solution just given is still shortest, and is one of four shortest solutions.
If a woman in the boat at the shore (but not on the shore) counts as being by herself (i.e. not in the presence of any men on the shore), then this puzzle can be solved in 9 one-way trips:
| Trip number | Starting bank | Travel | Ending bank |
|---|---|---|---|
| (start) | a b c | ||
| 1 | b c | a → | |
| 2 | b c | ← a | |
| 3 | c | ab → | |
| 4 | c | ← b | a |
| 5 | c | b → | a |
| 6 | c | ← b | a |
| 7 | bc → | a | |
| 8 | ← c | a b | |
| 9 | c → | a b | |
| (finish) | a b c |
Read more about this topic: Missionaries And Cannibals Problem
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