Giant, Slimy, Boy, Romy and Fatty want to cross a river.
To cross the river there is a small boat.
The boat only has 4 spaces. Every one can row the boat.

At the boat, Giant needs 4 spaces.
Slimy only needs 1 space, Boy also only needs 1 space.
Romy needs 2 spaces, and Fatty needs 3 spaces.

Slimy and Romy will fight to death if they left alone.
Slimy and Fatty also can not left alone.
Boy and Fatty can not left alone too.
("can not left alone" means both "at" and "out of" the boat).

How do they cross the river, with only 7 trips?

  • 1
    $\begingroup$ 1 trip means 1 way trip, NOT round trip. $\endgroup$ Commented Dec 7, 2016 at 7:55
  • 1
    $\begingroup$ Yeah think I got it. Giant was annoying $\endgroup$ Commented Dec 7, 2016 at 8:06

2 Answers 2


They can make it like this:

First trip:

Slimy, Boy and Romy row across. Boy and Romy get out.

Second trip:

Slimy rows back and gets out.

Third trip:

Fatty rows across and gets out.

Fourth trip:

Boy rows back and gets out.

Fifth trip:

Giant rows across and gets out

Sixth trip:

Romy rows back across

Last trip:

Romy, Boy and Slimy all row across

And they are all across the river with no-one dead or fighting.

How I got this:

First it's obvious that more than two people have to be across before Giant can make the journey because he fills the spaces. I also figured that the first had to be those three because otherwise you'd at some point be left with a fight-to-the-death scenario. This is why they all row across and then they each row back on their own after a while. The last trip is then the same as the first trip but the rest are already across the river. After that it was working out which order the rest crossed in, which took a bit of guesswork and trial and error. (I think) there's only one way of getting them all across (though @EspeciallyLime points out that the reverse of this also works)...

  • $\begingroup$ How do you answer it so fast ? $\endgroup$ Commented Dec 7, 2016 at 8:10
  • $\begingroup$ @JamalSenjaya Actually I'll add an explanation. $\endgroup$ Commented Dec 7, 2016 at 8:11

The reverse of Beastly Gerbil's solution also works:

1: B,S and R row across; B and S get out.
2: R rows back and gets out.
3: G rows across and gets out.
4: B rows back and gets out.
5: F rows across and gets out.
6: S rows back.
7: R, S and B row across.

These are the only two possibilities.

Note that Giant can't share a boat with anyone, and neither can Fatty (anyone who fits with him is someone he can't be alone with). So we need two forward trips there just to get them across. If we are to do the whole thing in seven trips that is four forward and three back. Each trip back decreases the number of people who are across by at least 1, two forward trips only contribute 1 and the others at most 3, and the total at the end must be 5. So the only possibility is for three people (who must be BSR) to cross in each of the other two forward trips, and one person (who must be one each of B,S,R) to cross back on each backward trip. So the two B,S,R trips must be the first (so that there is one of them to come back) and last (so that the last person to come back gets across). The first person to come back must be one of S and R (to avoid those being left alone) and likewise the last person to come back must be the other one (else they would have been alone immediately before). Once we choose what order S and R come back in, that determines everything except which order F and G cross in. Exactly one order of F and G will work for each order of S and R, to avoid F and S being left together.


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