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There are 3 men and 3 girls. lets name them ABC, A's daughter X, B's Daughter Y, C's daughter Z.

Rule to cross the river:

  1. At once Only two Can go on boat.
  2. Only Men know how to drive the boat (So there has to be atleast one man on boat)
  3. No girl crosses river with strange man, (that means, only father can take a girl on boat)
  4. No girl stays with strange men on shore, (that means, father should not leave his daughter with other men, either she should be alone, or with other girls or with her father)

How do they cross river?

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  • A and X cross the river, X stays, A rows back
  • B and Y cross the river, Y stays, B rows back
  • A and B cross the river, A stays, B and Y row back
  • B and C cross the river, B stays, C rows back
  • C and Z cross the river, both stay, B rows back
  • B and Y cross the river.
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  • $\begingroup$ Sounds like the Tower of Henoi puzzle. $\endgroup$ – Sukima Nov 3 '12 at 17:12
  • $\begingroup$ @Sukima, quite dissimilar isn't it :) $\endgroup$ – InfantPro'Aravind' Nov 4 '12 at 11:27
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I assume that the boat can only carry two persons, otherwise there is not much of a puzzle.

Martini has given the natural and obvious solution. The weakness of this solution is that in the second step, A can do nothing to prevent B from stepping ashore and ravish X.

The original version of this problem, posed by Alcuin of York (teacher of Charlemagne) more than 1200 years ago, is slightly different. There three men are travelling with their sisters, and each man desires the other girls. The question is how to cross the river without any girl being defiled. Martini's solution is not permitted, but the girls are able to row.

A mathematical method to solve problems of this kind is to look for paths in graphs. Each permissible distribution of people among the various places is a node in the graph, and two nodes are connected by an edge if it is possible to get from one node to the other with one boat trip. A solution of the problem is a path from a start node to an end node. There are simple and effective algorithms for finding paths in graphs, which makes it feasible to solve problems of this kind with hundreds of constraints.

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  • 3
    $\begingroup$ Another variant (Italian I think) relaxes constraint 3 to allow the case where the man won't ravish someone's sister if his sister is also present as a chaperone) $\endgroup$ – Foon Nov 3 '12 at 14:27
  • $\begingroup$ Corrected the question. $\endgroup$ – InfantPro'Aravind' Nov 4 '12 at 11:25
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Does a man in boat count as being next to a girl on shore? If so it is impossible to solve because if A and B go across C will be with X and Y. and if A and X go across A must come back and then only A can cross back which is not much help. If not then:
A+X->
A <-
B+Y->
B <-
A+B->
B+Y<-
B+C->
B<-
B+Y->
C<-
C+Z->

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1- one father nd his daughter crosses the river nd only father came back 2- then two girls go to other side nd a girl come back 3- then two other father go nd one girl with her father came back 4- then both father go nd the single girl came back 5- two girl go nd one girl came back 6- then two last girls goes.

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  • 2
    $\begingroup$ I think this violates Rule 2. $\endgroup$ – Ken Jun 17 '15 at 3:33
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[ A boatman wants to transport fox, goat and a grass bundle from one side of a river to another side using his boat. This Boatman can take maximum two of them at once with him on this boat... view fall : http://ynot.membrainsoft.com/2016/11/river-crossing-puzzle-ask-in-interviews.html

]1

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