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There are 3 cannibals and 3 missionaries on one side of a river

  1. Jul 5, 2014 #1
    In the textbook used for my problem solving and proofs course, one of the questions asks me to develop a geometric or algebraic representation of various questions. Here is one that I'm having trouble with:

    "Three cannibals and three missionaries are together on one side of the river. They wish to cross the river, and they have a boat that can carry two people. Describe a procedure for transporting all six people across the river in such a way that there are never more cannibals than missionaries on any one side at a given time."

    I tried really hard, and all I could come up with was this (don't laugh):

    1) let "a" be the number of missionaries on side 1
    2) "b" cannibals
    3) "c" missionaries side 2
    4) "d" cannibals

    a > b and c > d at all times.

    What am I doing wrong?
     
  2. jcsd
  3. Jul 5, 2014 #2
    1) The problem asks you to find a procedure that satisfies certain conditions/goals. You have merely expressed part of that goal using mathematical notation. You need to describe the procedure that satisfies that goal.
    2) The problem says "never more than", so the goal is ##a\geq b## and ##c\geq d## at all times.
    3) It doesn't look like you're taking into account the folks in the boat.

    Here is a link to a similar problem to give you an idea what your solution should look like. http://en.wikipedia.org/wiki/Fox,_goose_and_bag_of_beans_puzzle

    I'll trust that you won't be a cheater and click the link on the bottom of that page.
     
  4. Jul 6, 2014 #3

    verty

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    Hint: do you agree that it can never happen that one missionary is by himself on the boat?
     
  5. Jul 6, 2014 #4

    LCKurtz

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    Sure he can. A missionary can take one cannibal across and drop her off. Then he can go back and forth alone as much as he wants. He won't make much progress though. But I do have a question, can the cannibals be trusted to do what they are told, so you could send two cannibals across and have one return, for example?
     
  6. Jul 7, 2014 #5
    Wouldn't dropping off a cannibal mean that one side has more cannibals than missionaries?
     
  7. Jul 7, 2014 #6

    HallsofIvy

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    Which is not a problem if the number of missionaries on that side is 0!

    For example you could start by having one missionary and one cannibal cross. The missionary goes back alone (1 cannibal on one side, two cannibals and two missionaries on the other, one missionary in the boat.), picks up another cannibal, drops him off and goes back alone (two cannibals on one side, two missionaries and one cannibal on the other, one missionary in the boat). Now what must be done?
     
    Last edited: Jul 7, 2014
  8. Jul 7, 2014 #7

    LCKurtz

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    I was wondering if that's allowed. When he drops the second cannibal off there are momentarily 2 cannibals and 1 missionary on that side. Also as I asked earlier, can the cannibals be trusted to follow instructions? If so there is a solution avoiding even momentarily violating the cannibal/missionary ratio.
     
    Last edited: Jul 7, 2014
  9. Jul 7, 2014 #8

    HallsofIvy

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    That question occurred to me. I have this image of the single missionary picking up the single cannibal on the boat with him and throwing the cannibal the last few feet to the shore!
     
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