On Jealous Husbands and Confused Wives

Here we have a variation on the classic goat, wolf, cabbage, boat problem, courtesy of Sir Ian Stewart.

We have three extremely jealous and mistrustful husbands and three rather emotionally exhausted, hurt, and confused wives trying to get across a river. There is a boat with a set of oars but no oarsman, so they will have to row each other across. However, the boat can only hold two people.

Each husband is so incredibly suspicious of his wife that he does not trust his wife with any of the other men, even if their wives are present.

How can they all use the boat to get across the river without making any of the husbands incensed?

Alright, there are two easy solutions which involve multiple trips... You didn't specify the number of trips so I assumed unlimited.

We have:
Husband A (A) and Wife A (Aw)
Husband B (B) and Wife B (Bw)
Husband C (C) and Wife C (Cw)

Option One:
A & Aw, leave Aw and A travels back.
A Stays ashore.
B & Bw, leave Bw and B travels back.
B stays ashore.
C & Cw, leave Cw and C travels back.
Then left are only the husbands.
Say, A & B.. Leave A. B travels back.
Then B & C.. All ashore.

Or...

Option Two:
Aw & Bw, leave Aw and Bw travels back.
Bw & Cw, leave Bw and Cw travels back.
Cw & C, leave Cw and C travels back.
Then left are only the husbands.
Say, A & B.. Leave A. B travels back.
Then B & C.. All ashore.

Not the prettiest or most time-efficient of plans.. But the most obvious, for sure. lol

Excellent attempts Miss Anscombe!

However, I regret to inform you that your solutions are flawed (but clever!).

In Option One:

At the second step (or the third depending on how you denote steps) you have Aw, B, Bw all on the same shore. Since B and Aw are on the same shore without A present, husband A is quite incensed.

In Option Two:

Similarly you end up with all the wives on the far bank with husband C present with them, inflaming not only A but also B.

Regardless your answers remain clever and the single best solution for them would be marriage counseling.

Hahaha. Do tell the solution.. Please. Haha.

Well, I can give you a hint.

It'd be REALLY nice if all the husbands were on the other side of the river.

Another hint: the shortest possible set of moves is 11.

Husband A and Husband B go for a ride...
Leave Husband A.
Husband B and Husband C go for a ride...
Leave Husband B.
Husband C travels back and gets his own wife...
Husband C and his own wife BOTH get off the boat...
Husband B travels back and gets his own wife...
Husband B and his own wife BOTH get off the boat...
Husband A travels back and gets his own wife...
All are across the river...


Do-able? No wife is alone with a wife-less man... And no need to be jealous of the other husbands if their own husband is there as well, right?

If only; nevertheless the nature of the problem forces me to mention that I stated the following: "Each husband is so incredibly suspicious of his wife that he does not trust his wife with any of the other men, even if their wives are present."
It is also important to note that when A & B go for a ride, that leaves C with Aw, Bw, and Cw...

But you're thinking along the right track!

Here's a push in the right direction:
It'll be easier to get the husbands all on the destination side first if all the wives go first.
Thus your first move should be something along the lines of:
Aw & Cw
or
A & Aw

(There are no other possibilities since any other combination of two wives or a wife and her husband are congruent to any others mathematically speaking)

Sadly, I must give up.

In the case that you cannot have any wife with any other husband, regardless of companionship, I fear I have hit a brick wall. As soon as the first husband came of, jealousy would run rampant. I give up and surrender the thoughts and headaches to someone else. lol

Thanks for the mind-bender. (:

No problem, and I'm quite sad to see you give up!
I would encourage you to continue but for the headaches.

If it helps at all the way to prove that 11 is the shortest possible path (as well as to see all possible solutions) is a 6 dimensional hypercube where each vertex is a husband or wife.

Alright... You had a 21 year old college student sitting on the floor, playing with candy... Congratulations.

For sanity's sake, capitals are husbands and lowercases are wives.

1. A and a cross the river.
2. A comes back.
3. b and c cross the river.
4. a comes back.
5. B and C cross the river.
6. B and b come back.
7. A and B cross the river.
8. c comes back.
9. a and c cross the river.
10. B comes back.
11. B and b cross the river.

Yes?

Wheeeeeeeeeeeeeeeeeee!

Absolutely correct! Even better it is the minimum length possible solution!

And people said that playing with your food never got you anything.

This puzzle is now considered solved by Miss Liz Anscombe

Ahh! Yayy!