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Three missionaries on the left side of a river and three cannibals on the right side of the river want to cross a river using a boat which can carry at most two people, The boat is on the cannibals side. Under the constraint that, for both banks, if there are missionaries present on the bank, they cannot be outnumbered by cannibals (if they were, the cannibals would eat the missionaries). The boat cannot cross the river by itself with no people on board.

Plan minimum number of trips in such a way that all the cannibals go to the left side and missionaries come to the right side without any bloodshed.

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    $\begingroup$ Possible duplicate of Strategy to solve the Missionaries and Cannibals problem $\endgroup$ Nov 14 '17 at 12:56
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    $\begingroup$ Not a duplicate - the usual formulation (and that used in the putative duplicate) has all 6 people starting on one side and needing to cross to the other side without bloodshed. This problem has them on opposite banks and needing to trade places. $\endgroup$
    – Rubio
    Nov 14 '17 at 14:09
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It should take 5 crossings:

MMM <C>CC 1C crosses left
MC<MM> CC 2M cross right
MC <MC>MC 1M 1C cross left
CC<MM> MC 2M cross right
CC <C>MMM 1C crosses left
CCC<> MMM

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  • $\begingroup$ on a lighter vein, does JS1 stand for Jamal Senjaya One ? $\endgroup$ Nov 14 '17 at 9:40
  • $\begingroup$ @MeaCulpaNay No, just a coincidence I guess. $\endgroup$
    – JS1
    Nov 14 '17 at 20:47
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Actually it should be doable in 4 crossings, since the question calls for the Cannibals not outnumbering the Missionaries, so I assume same amount of Cannibals and Missionaries is okay and embarking and disembarking is instantaneous, so no snatching of boarding missionaries while you reunite with your cannibal friends.
MMM----------[CC]C 2C cross left
MMCC[M]----------C 1M crosses right
MMCC----------[C]MM 1C crosses left
CCC[MM]----------M 2M cross right

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  • $\begingroup$ Are there four cannibals in the middle two steps? I'm not sure how two C cross left and leave 2 behind, since there are only supposed to be 3 C. Also, aren't there 4 crossings, not 3, in your solution? $\endgroup$
    – bobble
    May 8 at 18:22
  • $\begingroup$ Thanks, you're right. I was struggling with the editing a bit. $\endgroup$
    – Mookuh
    May 8 at 18:24
  • $\begingroup$ I still see four cannibals in the second step? Two on either side. $\endgroup$
    – bobble
    May 8 at 18:26
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    $\begingroup$ I wasn't done struggling D: $\endgroup$
    – Mookuh
    May 8 at 18:26
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    $\begingroup$ Final nitpick... third step has two M on either side? $\endgroup$
    – bobble
    May 8 at 18:31

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