3 couples on one side of a river. Aa Bb Cc (caps representing husbands, lowercase representing wives)

One boat in which 2 people can travel at a time.

At least one person must row the boat wherever it needs to be taken. The boat cannot travel to any side by water current.

People can travel with someone of the other gender only if they are their spouse. No man or woman can travel with someone they are not married with. (This is what makes this puzzle different from $n$ couples crossing a river)

No woman can be left on a side with another man unless her husband is also present there.

What is the method for all of them crossing the river?

  • $\begingroup$ You say "No woman can be left on a side with another man". What if there's a brief stopover? Like what if a is on the far side alone, B and b row over, then B comes right back. Is that prohibited? $\endgroup$
    – Duncan
    Commented Mar 4, 2016 at 22:52
  • $\begingroup$ Yes. It is prohibited. $\endgroup$ Commented Mar 4, 2016 at 22:54
  • 1
    $\begingroup$ I'm not seeing how it's different from the linked question. Can you give an example of something prohibited in this problem that is allowed in the other one? $\endgroup$
    – f''
    Commented Mar 4, 2016 at 23:05
  • $\begingroup$ Not allowing the travelling part. And not allowing any person to move to the other bank with opposite gender person(s) even for a touchdown. $\endgroup$ Commented Mar 4, 2016 at 23:16
  • 3
    $\begingroup$ Wife not travelling with anyone of the other gender but your spouse is part of the original - this is basically the same puzzle except that asked for the formula (but many answers show the pattern as well). Unless you can point to some part that people are miss-understanding. $\endgroup$ Commented Mar 5, 2016 at 0:10

2 Answers 2


Here goes:

Aa cross over. A crosses back.
Departure side: A, Bb, Cc
Destination side: a

Next, bc row over, c rows back alone.
Departure side: A, B, Cc
Destination side: ab

AB row over together, Bb row back.
Departure side: Bb, Cc
Destination side: Aa

Now the two men head over. BC to destination. Send a back.
Departure side: abc
Destination side: ABC

Now it's simple. ab over, a back, then ac over and done!


I'll solve it.

$[-,-,-,-,-,-]$ $\leftarrow (a,b)$ $[A,a,B,b,C,c]$

$[a,b,-,-,-,-]$ $\rightarrow (b)$ $[A,B,C,c,-,-]$

$[a,-,-,-,-,-]$ $\leftarrow (b,c)$ $[A,a,B,b,C,-]$

$[a,b,c,-,-,-]$ $\rightarrow (c)$ $[A,B,C,-,-,-]$

$[a,b,-,-,-,-]$ $\leftarrow (A,B)$ $[A,B,C,c,-,-]$

$[a,A,b,B]$ $\rightarrow (a,A)$ $[C,c,-,-,-,-]$

$[b,B,-,-,-,-]$ $\leftarrow (A,C)$ $[C,c,A,a,-,-]$

$[A,C,B,b]$ $\rightarrow (b)$ $[a,c,-,-,-,-]$

$[A,B,C,-,-,-]$ $\leftarrow (b,c)$ $[a,b,c,-,-,-]$

$[A,B,b,C,c,-,-]$ $\rightarrow (b)$ $[a,-,-,-,-,-]$

$[A,B,C,c,-,-]$ $\leftarrow (b,a)$ $[a,b,-,-,-,-]$

$[A,a,B,b,C,c]$ $[-,-,-,-,-,-]$

$11$ Trips total.
$6$ forwards.
$5$ backwards.

  • $\begingroup$ Someone please help me format my answer correctly. $\endgroup$ Commented Dec 6, 2016 at 1:06
  • $\begingroup$ @Duncan, I think my method is faster than yours? $\endgroup$ Commented Dec 6, 2016 at 1:13
  • 1
    $\begingroup$ This answer is essentially identical to Duncan's. It follows the same basic pattern of males or females or couples moving, and takes the same number of trips. Duncan's answer was marked accepted 9 months ago; it takes only basic counting to verify your answer is not faster than Duncan's, and a quick skimming to see that the answers are fundamentally equivalent. I'm glad you're eager to contribute, but answering old questions which have already had an answer accepted is unlikely to get you any positive notice. Try answering a current puzzle! $\endgroup$
    – Rubio
    Commented Dec 6, 2016 at 1:26
  • $\begingroup$ OK. I won't necro old questions. $\endgroup$ Commented Dec 6, 2016 at 1:43

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