Alexander Shapovalov Crosses a River

Alexander Shapovalov is a prolific puzzle writer. He has a special webpage of his river-crossing puzzles (in Russian). Here is one of these puzzles.

Three swindlers have two suitcases each. They approach a river they wish to cross. There is one boat that can carry three objects, where a person or a suitcase counts as one object. No swindler can trust his suitcase to his swindler friends when he is away, but each swindler doesn’t mind his suitcases left alone at the river shore. Can they cross the river?

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10 Comments

  1. Leo B.:

    Two of them approach the river from one shore, and the third – from the other?

  2. tanyakh:

    Leo,

    The puzzle assumes that they approach from the same shore.

  3. Calcdave:

    I think I have a solution

  4. Jonathan Kariv:

    Spoiler alert:

    Swindler A takes his across and leaves them there. He then returns alone (no suitcases) to the near side. Next all 3 cross together and B and C return. Leaving A on the far side with his suitcases and B and C on the near side with the boat and their cases. B then takes the boat and his two cases across. Leaving A, B and there stuff on the far side with the boat and C on the near with his cases. A and B return together and fetch C. C then takes the boat back alone to collect his stuff.

  5. Leo R:

    Can a boat with no people cross the river?

  6. L33tminion:

    I have a solution, in seven crossings.

  7. Per:

    Haha, the math world is so small! I remember going to Alexanders math circle in Stockholm to do problem solving.

  8. Vincent:

    At least some of Shapovalov’s puzzles are now also available in English: http://ashap.info/Zadachi/eng/River.html

  9. armadillozenith:

    Yes, they (the swindlers) can cross the river! They all simply get into the boat and go across together in one trip.
    The puzzle does not state that their cases have to go with them. It specifically says that “each swindler doesn’t mind his suitcases left alone at the river shore”.

  10. Shakir:

    Jonathan Kariv’s answer is correct

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